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✓ 19 results · all verified · 18 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Absolute and Conditional Convergence; Rearrangement; Products

1 · Prerequisites

2 · Summary

A note on the notation $ι$ . A natural number here is a von Neumann natural, that is a set, so it is not an element of $R$ and cannot be divided into $1$ . The canonical natural $ι (n) = n \cdot 1_{R}$ is the real number that $n$ names (Canonical naturals are positive and strictly increasing), so $1 / ι (k + 1)$ is what an informal text writes as $1 / (k + 1)$ ; the shift by one is there because $N$ contains $0$ and $ι (0) = 0$ .

Objective. The previous page decided whether a series converges. This page asks what a convergent series is worth as an object: may its terms be reordered, may two such series be multiplied, may brackets be inserted or removed, may a doubly indexed family be summed in either order. The answer turns out to depend on a single dividing line, drawn in the first item of the page, and the whole page is the story of that line.

The dividing line. Absolutely convergent and conditionally convergent series, and the general starting index calls $\sum a_{k}$ absolutely convergent when $\sum ∣ a_{k} ∣$ converges and conditionally convergent when it converges without that. One implication is already proved on the previous page and is not restated here: If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges says absolute convergence implies convergence, which is what makes the two words partition the convergent series. The converse fails, and FALSE: every convergent series converges absolutely exhibits the alternating harmonic series as the witness. Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ is the technical form of the distinction: for an absolutely convergent series both part series $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent one both diverge to $+ \infty$ . The rearrangement and unconditional-convergence results, together with the Cauchy-product and double-series results, are organised by that dichotomy. The three convergence tests that come first instead arise from summation by parts, while the grouping, infinite-product, and decimal results have their own hypotheses.

Two convergence tests that need no sign pattern. Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ is the discrete integration by parts, $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for $n \geq 1$ . From it Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges follows at once: bounded partial sums of $\sum a_{k}$ together with a nonincreasing null $(b_{k})$ give convergence of $\sum a_{k} b_{k}$ . The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ is the special case with the alternating sequence, and it carries in addition the bracketing of the sum between consecutive partial sums and the error bound $∣ L - t_{n} ∣ \leq b_{n}$ , which the Dirichlet estimate does not produce and which is proved here from the interlacing of the even-index and odd-index partial sums. Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges trades the two hypotheses: a convergent $\sum a_{k}$ against a monotone bounded factor.

Rearrangement. Rearrangement of a series along a bijection of $N$ , and unconditional convergence fixes the vocabulary: a rearrangement is a composite with a bijection of $N$ , and unconditional convergence means every rearrangement converges to the same sum. Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum proves that absolute convergence suffices, and the proof reduces everything to the nonnegative case, where the sum is a supremum and cannot see the order at all. The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ is the opposite extreme: for a conditionally convergent real series and any $α \leq β$ in the extended reals there is a rearrangement whose partial sums have limit inferior $α$ and limit superior $β$ . In particular every real is the sum of some rearrangement, and rearrangements diverging to $+ \infty$ and to $- \infty$ exist. For a series of real numbers, unconditional convergence and absolute convergence are the same property closes the circle: over $R$ , absolute convergence, unconditional convergence, and the mere requirement that every rearrangement converge, are the same property. FALSE: every rearrangement of a convergent series converges, and to the same sum records the naive claim these results refute, and The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order says what the same question looks like for series of vectors and why that answer is not reachable at this point in the reading order.

Brackets. Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum shows that a convergent series may be grouped into blocks at will, the grouped partial sums being a subsequence of the original ones. The converse fails, and FALSE: if some grouping of a series converges then the series itself converges gives the reason: a subsequence of a divergent sequence may converge.

Products. The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ fixes $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ , the coefficients forced by multiplying two power series. Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ proves that one factor converging absolutely and the other merely converging already give $\sum c_{n} = A B$ ; its first claim is a finite identity, $\sum_{n < N} c_{n} = \sum_{i < N} a_{i} B_{N - i}$ , holding for arbitrary sequences, and that identity is reused for the absolute values in If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ , where both factors are absolutely convergent and the product is too. FALSE: the Cauchy product of two convergent series converges shows that convergence of both factors alone is not enough.

Double series. Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value proves that when each row of an array is absolutely summable and the row totals are summable, the two iterated sums and the sum along every bijection $N \to N \times N$ all exist and agree. Independence of the enumeration is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum again. Without the absolute hypothesis the two iterated sums can both exist and differ, which is FALSE: whenever both iterated sums of a double array exist, they are equal.

Infinite products. Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors defines convergence of $\prod a_{k}$ as convergence of some tail of the partial products to a nonzero limit, and says why a zero limit has to be excluded. For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent carries the elementary theory: the Weierstrass bounds $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k})$ , with the companion upper bound $\prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ whenever $\sum_{k < n} p_{k} < 1$ , the equivalence of $\prod (1 + p_{k})$ with $\sum p_{k}$ for nonnegative terms, the $(1 - p_{k})$ form together with the fact that its partial products tend to $0$ when $\sum p_{k}$ diverges, and convergence of $\prod (1 + p_{k})$ from convergence of $\sum ∣ p_{k} ∣$ for signed terms. FALSE: $\prod (1 + p_{k})$ converges whenever $p_{k} \to 0$ records that factors tending to $1$ decide nothing. No logarithm is used anywhere on this page; every one of those inequalities is an induction on finite products, and the refinement usually proved with logarithms is deferred, as Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for records.

Decimal expansions. Base- $b$ expansions: for an integer $b \geq 2$ every $x \in [0, 1)$ is the sum of $\sum_{j \geq 0} d_{j} / b^{j + 1}$ for digits $d_{j} < b$ , and the digit sequence is unique among those that are not eventually constantly $b - 1$ is the payoff of the geometric series in this direction: for an integer $b \geq 2$ every $x \in [0, 1)$ is the sum of $\sum_{j \geq 0} ι (d_{j}) / β^{j + 1}$ for a digit sequence that is unique once the sequences eventually constantly $b - 1$ are excluded. The construction is floor-free: the integer part of a real is developed later in the reading order, so the digit at each stage is selected by a finite case distinction closed by the well-ordering principle, and the digits are assembled by the recursion theorem.

What is proved and what is only proved to exist. Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for goes through the page and says which sums are named without being evaluated (the alternating harmonic sum, the sum of its two-positive-one-negative rearrangement) and what each evaluation waits for. Every scope statement on this page is relative to the reading order: the material named is developed elsewhere in this library, later than this page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Absolutely convergent and conditionally convergent series, and the general starting index

Definition

Let $(a_{k})$ be a sequence of reals, with series $\sum a_{k}$ and partial sums $s_{n} = \sum_{k < n} a_{k}$ as in Series, partial sums, convergence and the sum, divergence, and the tail series, and let $∣ x ∣$ be the absolute value (Absolute value in an ordered field).

Absolute convergence. The series $\sum a_{k}$ converges absolutely when the series $\sum ∣ a_{k} ∣$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series). Since $∣ a_{k} ∣ \geq 0$ for every $k$ (Basic properties of the absolute value), this is a statement about a series of nonnegative terms.

Conditional convergence. The series $\sum a_{k}$ converges conditionally when it converges (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals) and does not converge absolutely.

So a convergent series is exactly one of the two: absolutely convergent or conditionally convergent, according as $\sum ∣ a_{k} ∣$ converges or not.

One implication is already proved, and is not reproved anywhere on this page. If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges states that if $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges. That lemma was coined and proved on the previous page of this track, where the root and ratio tests need it; this page names it and builds on it. In particular an absolutely convergent series is a convergent series, so the two words above really do partition the convergent series, and "conditionally convergent" is not vacuous by accident: the alternating harmonic series is a witness, and the witness is exhibited in FALSE: every convergent series converges absolutely.

General starting index. Let $m \in N$ and let $(a_{k})_{k \geq m}$ be a family from $m$ (Series, partial sums, convergence and the sum, divergence, and the tail series). The series $\sum_{k \geq m} a_{k}$ converges absolutely when $\sum_{k \geq m} ∣ a_{k} ∣$ converges, and converges conditionally when it converges and does not converge absolutely. By Series, partial sums, convergence and the sum, divergence, and the tail series both statements are the corresponding statements for the shifted sequence $j \mapsto a_{j + m}$ , so nothing new is being defined and every result below transfers to a general starting index in the same way, exactly as If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges already records for the one implication it proves.

Remarks

Absolute convergence is a condition on the terms, not on the sum. It says the series of absolute values converges, and it says nothing about the value of $\sum_{k = 0}^{\infty} a_{k}$ . The two sums are in general different, and no statement here identifies them.
Why the distinction earns a page. Every result on this page separates the two classes. An absolutely convergent series may be reordered at will (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum) and multiplied by another (Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ ); a conditionally convergent one may be reordered to any sum whatever (The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ ). The difference is not one of degree.
A series of nonnegative terms converges absolutely if it converges at all, since then $∣ a_{k} ∣ = a_{k}$ . So the distinction is invisible for the comparison, condensation, Raabe, Gauss and Kummer tests of the previous page, all of which assume terms of one sign. It is not invisible on that page as a whole: the root and ratio tests are stated for terms of arbitrary sign and reach convergence of $\sum a_{k}$ precisely through If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, which is where the word absolutely convergent is first used. What that page does not develop, and this one does, is everything that separates the two classes rather than the one implication those two tests need.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$

Statement

Let $(a_{k})$ be a sequence of reals (Series, partial sums, convergence and the sum, divergence, and the tail series) and define its positive part and negative part by

$a_{k}^{+} : = \frac{∣ a_{k} ∣ + a_{k}}{2}, a_{k}^{-} : = \frac{∣ a_{k} ∣ - a_{k}}{2} (k \in N),$

with $∣ x ∣$ the absolute value (Absolute value in an ordered field). Then:

$a_{k}^{+} = \max {a_{k}, 0}$ and $a_{k}^{-} = \max {- a_{k}, 0}$ (Maximum and minimum of a set); in particular $a_{k}^{+} \geq 0$ and $a_{k}^{-} \geq 0$ , and $a_{k} = a_{k}^{+} - a_{k}^{-}, ∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-} .$
$\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index) if and only if both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge.
If $\sum a_{k}$ converges conditionally, then neither $\sum a_{k}^{+}$ nor $\sum a_{k}^{-}$ converges, and the partial sums of each diverge to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ).

Claim 3 is the engine of the rearrangement theory: a conditionally convergent series carries an unlimited supply of positive terms and an unlimited supply of negative ones, and its convergence is nothing but a cancellation between them.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals, its positive and negative parts $a_{k}^{+}$ and $a_{k}^{-}$ as displayed above, and the partial sums of the associated series (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

Absolute value: $∣ x ∣ \geq 0$ , $- ∣ x ∣ \leq x \leq ∣ x ∣$ , and $∣ x ∣ = x$ when $x \geq 0$ while $∣ x ∣ = - x$ when $x < 0$ (Absolute value in an ordered field, Basic properties of the absolute value).

[L2]

A maximum of a subset of $R$ is its greatest element, and there is at most one (Maximum and minimum of a set).

[L3]

Linearity of series: if $\sum x_{k}$ and $\sum y_{k}$ converge then so does $\sum (x_{k} + y_{k})$ , and $\sum c x_{k}$ converges for every real $c$ (Convergent series add and scale termwise).

[L4]

Direct comparison: if $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum y_{k}$ converges, then $\sum x_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L5]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above; and if that range is not bounded above then the partial sums diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Divergence to $+ \infty$ and to $- \infty$ ).

[L6]

$\sum a_{k}$ converges absolutely means $\sum ∣ a_{k} ∣$ converges, and converges conditionally means it converges while $\sum ∣ a_{k} ∣$ does not (Absolutely convergent and conditionally convergent series, and the general starting index, Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

For every $k$ , $∣ a_{k} ∣ + a_{k} \geq 0$ and $∣ a_{k} ∣ - a_{k} \geq 0$ , since $- ∣ a_{k} ∣ \leq a_{k} \leq ∣ a_{k} ∣$ ; dividing by the positive real $2$ gives $a_{k}^{+} \geq 0$ and $a_{k}^{-} \geq 0$ .

L1algebra

1.2

For every $k$ , $a_{k}^{+} - a_{k}^{-} = ((∣ a_{k} ∣ + a_{k}) - (∣ a_{k} ∣ - a_{k})) / 2 = a_{k}$ and $a_{k}^{+} + a_{k}^{-} = ((∣ a_{k} ∣ + a_{k}) + (∣ a_{k} ∣ - a_{k})) / 2 = ∣ a_{k} ∣$ .

algebra

1.3

Assume now that $\sum a_{k}$ converges conditionally, so $\sum a_{k}$ converges and $\sum ∣ a_{k} ∣$ diverges.

2.1

If $a_{k} \geq 0$ then $∣ a_{k} ∣ = a_{k}$ , so $a_{k}^{+} = a_{k}$ and $a_{k}^{-} = 0$ ; if $a_{k} < 0$ then $∣ a_{k} ∣ = - a_{k}$ , so $a_{k}^{+} = 0$ and $a_{k}^{-} = - a_{k}$ . In both situations $a_{k}^{+}$ is the greater of $a_{k}$ and $0$ and $a_{k}^{-}$ is the greater of $- a_{k}$ and $0$ , which is claim 1 together with step 1.1 and step 1.2.

L1L2step 1.1step 1.2algebra

2.2

From step 1.1 and step 1.2, $0 \leq a_{k}^{+} \leq a_{k}^{+} + a_{k}^{-} = ∣ a_{k} ∣$ and $0 \leq a_{k}^{-} \leq ∣ a_{k} ∣$ for every $k$ .

step 1.1step 1.2algebra

2.3

If both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, then $\sum ∣ a_{k} ∣ = \sum (a_{k}^{+} + a_{k}^{-})$ converges.

step 1.2L3

3.1

If $\sum ∣ a_{k} ∣$ converges then, by comparison with $\sum ∣ a_{k} ∣$ using step 2.2, both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge.

step 2.2L4

3.2

If $\sum a_{k}^{+}$ converged, then $\sum a_{k}^{-} = \sum (a_{k}^{+} + (- 1) a_{k})$ would converge by linearity, whence $\sum ∣ a_{k} ∣$ would converge by step 2.3; since $\sum ∣ a_{k} ∣$ diverges, $\sum a_{k}^{+}$ diverges.

step 1.3step 1.2step 2.3L3

3.3

If $\sum a_{k}^{-}$ converged, then $\sum a_{k}^{+} = \sum (a_{k}^{-} + a_{k})$ would converge by linearity, whence again $\sum ∣ a_{k} ∣$ would converge; since $\sum ∣ a_{k} ∣$ diverges, $\sum a_{k}^{-}$ diverges.

step 1.3step 1.2step 2.3L3

4.1

Claim 2 is the conjunction of step 2.3 and step 3.1, read through the definition of absolute convergence.

step 2.3step 3.1L6

5.1

Both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ are series of nonnegative terms by step 1.1, so each diverges only if the range of its partial sums fails to be bounded above, and then those partial sums diverge to $+ \infty$ ; this is claim 3.

step 1.1step 3.2step 3.3L5∎

Remarks

The two parts are determined by the terms, with no choice anywhere. The displayed formulas define $a^{+}$ and $a^{-}$ outright, and step 2.1 identifies them with the two maxima; nothing in the proof selects one of several candidates.
Claim 3 is sharp in both directions. Absolute convergence makes both part series converge, and then $\sum_{k = 0}^{\infty} a_{k}$ is the difference of their sums. Conditional convergence makes both part series diverge to $+ \infty$ , and the difference of their partial sums is what converges. There is no third possibility for a convergent series, because claim 2 covers the case where one of them converges: if exactly one converged, $\sum a_{k} = \sum (a_{k}^{+} - a_{k}^{-})$ could not converge, since the sum of a convergent and a divergent series diverges.
Why $\max$ is mentioned at all. The formulas with $∣ a_{k} ∣$ are what the algebra uses, while $\max {a_{k}, 0}$ is what the name "positive part" means and what makes claims about signs immediate. Step 2.1 records that they agree, so either may be used later without further comment.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals and let

$A_{n} : = \sum_{k < n} a_{k} (n \in N)$

be the partial sums of $\sum a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion), so that $A_{0} = 0$ and $a_{k} = A_{k + 1} - A_{k}$ for every $k$ . Then for every natural number $n \geq 1$

$\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k}) .$

Both sides are finite sums in the sense of Finite sums and finite products, by recursion; at $n = 1$ the right-hand sum is empty and the identity reads $a_{0} b_{0} = A_{1} b_{0}$ .

The hypothesis $n \geq 1$ is what makes the statement legitimate, not merely convenient: the index $n - 1$ occurs on the right, and $n - 1$ is a natural number exactly when $n \geq 1$ . At $n = 0$ there is nothing to state, both the left-hand side and $A_{0}$ being $0$ .

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ of reals and the partial sums $A_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

Finite sums are defined by the recursion $\sum_{k < 0} x_{k} = 0$ and $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ (Finite sums and finite products, by recursion).

[L2]

The partial sums satisfy $A_{0} = 0$ and $A_{n + 1} = A_{n} + a_{n}$ for every $n$ , those being the two clauses of [L1] applied to $(a_{k})$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L3]

Finite sums are additive and may be split at any intermediate index (Laws of finite sums and finite products).

[L4]

The principle of induction on $N$ (The principle of mathematical induction).

Proof

technique · induction

1.1

The claim to be proved by induction is the statement $P (m)$ : the displayed identity holds at $n = m + 1$ , that is $\sum_{k < m + 1} a_{k} b_{k} = A_{m + 1} b_{m} - \sum_{k < m} A_{k + 1} (b_{k + 1} - b_{k})$ . Every $n \geq 1$ is $m + 1$ for exactly one $m \in N$ , so proving $P (m)$ for all $m$ proves the lemma.

1.2

$P (0)$ holds: the left-hand side is $\sum_{k < 1} a_{k} b_{k} = a_{0} b_{0}$ by [L1], while $A_{1} = A_{0} + a_{0} = a_{0}$ by [L2] and $\sum_{k < 0} A_{k + 1} (b_{k + 1} - b_{k}) = 0$ by [L1], so the right-hand side is $a_{0} b_{0} - 0$ .

L1L2base

1.3

Assume $P (m)$ for a fixed $m \in N$ .

1.4

By [L1], $\sum_{k < m + 2} a_{k} b_{k} = \sum_{k < m + 1} a_{k} b_{k} + a_{m + 1} b_{m + 1}$ .

1.5

By [L1], $\sum_{k < m + 1} A_{k + 1} (b_{k + 1} - b_{k}) = \sum_{k < m} A_{k + 1} (b_{k + 1} - b_{k}) + A_{m + 1} (b_{m + 1} - b_{m})$ .

L1L3

1.6

By [L2], $A_{m + 2} = A_{m + 1} + a_{m + 1}$ , so $a_{m + 1} = A_{m + 2} - A_{m + 1}$ .

2.1

Substituting the induction hypothesis into step 1.4 gives $\sum_{k < m + 2} a_{k} b_{k} = A_{m + 1} b_{m} - \sum_{k < m} A_{k + 1} (b_{k + 1} - b_{k}) + a_{m + 1} b_{m + 1}$ .

step 1.3step 1.4

2.2

Using step 1.6, $A_{m + 1} b_{m} + a_{m + 1} b_{m + 1} = A_{m + 1} b_{m} + A_{m + 2} b_{m + 1} - A_{m + 1} b_{m + 1} = A_{m + 2} b_{m + 1} - A_{m + 1} (b_{m + 1} - b_{m})$ .

step 1.6algebra

3.1

Combining step 2.1 and step 2.2 and then step 1.5 gives $\sum_{k < m + 2} a_{k} b_{k} = A_{m + 2} b_{m + 1} - A_{m + 1} (b_{m + 1} - b_{m}) - \sum_{k < m} A_{k + 1} (b_{k + 1} - b_{k}) = A_{m + 2} b_{m + 1} - \sum_{k < m + 1} A_{k + 1} (b_{k + 1} - b_{k})$ , which is $P (m + 1)$ .

step 2.1step 2.2step 1.5algebra

4.1

By [L4] applied to step 1.2 and step 3.1, $P (m)$ holds for every $m \in N$ , that is, the displayed identity holds for every $n \geq 1$ .

step 1.2step 3.1L4discharge-induction∎

Remarks

What the identity is for. It converts a series $\sum a_{k} b_{k}$ , about which nothing is assumed, into a boundary term $A_{n} b_{n - 1}$ and a series $\sum A_{k + 1} (b_{k + 1} - b_{k})$ whose terms carry the differences of $(b_{k})$ . If $(A_{n})$ is bounded and $(b_{k})$ is monotone, those differences have one sign and telescope, which is exactly the situation of Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges. The transformation is the discrete analogue of integration by parts, and the boundary term is the analogue of the boundary term there.
The block form needs no separate proof. For $1 \leq M \leq n$ , subtracting the identity at $M$ from the identity at $n$ gives $\sum_{k = M}^{n - 1} a_{k} b_{k} = A_{n} b_{n - 1} - A_{M} b_{M - 1} - \sum_{k = M - 1}^{n - 2} A_{k + 1} (b_{k + 1} - b_{k})$ , using only splitting of finite sums (Laws of finite sums and finite products). Nothing on this page needs that form, so it is recorded here rather than stated as a result.
Two conventions are doing work. $A_{n}$ sums the $n$ terms $a_{0}, \dots, a_{n - 1}$ , so $A_{0} = 0$ and $a_{k} = A_{k + 1} - A_{k}$ with no shift (Series, partial sums, convergence and the sum, divergence, and the tail series); and the empty sum is $0$ (Finite sums and finite products, by recursion), which is what makes $n = 1$ a genuine instance of the identity rather than a case to be excluded.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals, and let $A_{n} = \sum_{k < n} a_{k}$ be the partial sums of $\sum a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Suppose that

the range ${A_{n} : n \in N}$ is bounded (Lower bound, bounded below, bounded set), that is there is a real $M \geq 0$ with $∣ A_{n} ∣ \leq M$ for every $n$ ; and
$(b_{k})$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals).

Then $\sum a_{k} b_{k}$ converges.

Under hypothesis 2 the terms $b_{k}$ are automatically nonnegative, and the proof says so before using it: a nonincreasing sequence is bounded below by each of its own later terms, and passing to the limit gives $b_{k} \geq 0$ (Limits preserve non-strict inequalities).

Nothing is assumed about $\sum a_{k}$ itself. Its partial sums need only stay bounded; they need not converge. That is what makes this test the source of the alternating series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ ) and of examples whose sign pattern is not alternating at all.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ of reals with $A_{n} = \sum_{k < n} a_{k}$ bounded in absolute value, and $(b_{k})$ nonincreasing with $b_{k} \to 0$ .

[L1]

Abel summation by parts: for every $n \geq 1$ , $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ (Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ ).

[L2]

Nonincreasing means $b_{j} \geq b_{k}$ whenever $j \leq k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L4]

Telescoping: with $d_{k} : = b_{k} - b_{k + 1}$ , the partial sums of $\sum d_{k}$ are $b_{0} - b_{n}$ , and $\sum d_{k}$ converges if and only if $(b_{k})$ converges, with sum $b_{0} - \lim_{k} b_{k}$ ( $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ ).

[L5]

[L6]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[L7]

Linearity: if $\sum x_{k}$ converges then so does $\sum c x_{k}$ for every real $c$ (Convergent series add and scale termwise).

[L8]

A null sequence times a bounded sequence is null (A null sequence times a bounded sequence is null).

[L9]

Algebra of limits for differences of convergent sequences (Algebra of limits: sums, scalar multiples, products and quotients).

[L10]

A sequence converges to $x$ if and only if some tail of it converges to $x$ (Convergence depends only on the tail).

[L11]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , and $∣ - x ∣ = ∣ x ∣$ (Basic properties of the absolute value).

[L12]

A bounded set of reals admits a bound in absolute value (Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

Fix a real $M \geq 0$ with $∣ A_{n} ∣ \leq M$ for every $n \in N$ .

givenL12choose

1.2

For each fixed $k$ the inequality $b_{m} \leq b_{k}$ holds for all $m \geq k$ , and $(b_{m})_{m}$ converges to $0$ while the constant sequence with value $b_{k}$ converges to $b_{k}$ ; hence $0 \leq b_{k}$ .

givenL2L3

1.3

Put $d_{k} : = b_{k} - b_{k + 1}$ and $c_{k} : = A_{k + 1} (b_{k + 1} - b_{k})$ for $k \in N$ , and let $s_{n} : = \sum_{k < n} a_{k} b_{k}$ , $t_{n} : = \sum_{k < n} c_{k}$ and $u_{n} : = A_{n + 1} b_{n}$ .

given

1.4

Each $d_{k} \geq 0$ , since $(b_{k})$ is nonincreasing; and $\sum d_{k}$ converges, with sum $b_{0} - 0 = b_{0}$ , because $(b_{k})$ converges to $0$ .

givenL2L4

2.1

For every $k$ , $∣ c_{k} ∣ = ∣ A_{k + 1} ∣ ∣ b_{k + 1} - b_{k} ∣ = ∣ A_{k + 1} ∣ d_{k} \leq M d_{k}$ , using $b_{k + 1} - b_{k} = - d_{k}$ and $d_{k} \geq 0$ .

step 1.1step 1.3step 1.4L11

2.2

The sequence $(A_{n + 1})_{n}$ is bounded by $M$ and $(b_{n})$ converges to $0$ , so $u_{n} = A_{n + 1} b_{n}$ converges to $0$ .

step 1.1step 1.3givenL8

2.3

The series $\sum M d_{k}$ converges, by step 1.4 and linearity.

step 1.4L7

2.4

For every $n \in N$ , applying [L1] at the index $n + 1 \geq 1$ gives $s_{n + 1} = A_{n + 1} b_{n} - \sum_{k < n} A_{k + 1} (b_{k + 1} - b_{k}) = u_{n} - t_{n}$ .

step 1.3L1

3.1

Since $0 \leq ∣ c_{k} ∣ \leq M d_{k}$ for every $k$ , the series $\sum ∣ c_{k} ∣$ converges by comparison, and therefore $\sum c_{k}$ converges; write $T$ for its sum, so that $t_{n} \to T$ .

step 2.1step 2.3L5L6

4.1

By step 2.2, step 3.1 and the algebra of limits, $s_{n + 1} \to 0 - T = - T$ as $n \to \infty$ .

step 2.2step 3.1step 2.4L9

5.1

The sequence $(s_{n + 1})_{n \in N}$ is the first tail of $(s_{n})$ , so $(s_{n})$ itself converges to $- T$ ; that is, $\sum a_{k} b_{k}$ converges, with sum $- T$ .

step 4.1L10∎

Remarks

Where each hypothesis is used, and none is decorative. Boundedness of $(A_{n})$ is used twice: once to bound $∣ c_{k} ∣$ in step 2.1, and once to kill the boundary term in step 2.2. Monotonicity of $(b_{k})$ is what makes $∣ b_{k + 1} - b_{k} ∣$ equal to $b_{k} - b_{k + 1}$ , so that the bound in step 2.1 telescopes; without it the differences need not sum to anything. And $b_{k} \to 0$ is used both in the telescoping sum of step 1.4 and in the boundary term of step 2.2.
Why nonincreasing and not monotone, although either would do. Hypothesis 2 could equally be stated with "monotone", and the theorem would still be true: a nondecreasing $(b_{k})$ converging to $0$ is nonpositive, so $(- b_{k})$ is nonincreasing and converges to $0$ , and applying the theorem to it gives convergence of $\sum a_{k} (- b_{k})$ and hence of $\sum a_{k} b_{k}$ (Convergent series add and scale termwise). What "monotone" may not be weakened to is "monotone and bounded": a monotone $(b_{k})$ with a nonzero limit is not covered, and for such a factor the conclusion fails in general. The nonincreasing form is chosen here because it is the form the proof uses, and because it makes $b_{k} \geq 0$ immediate. Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges is the result that handles monotone bounded factors, and it has a different hypothesis on $\sum a_{k}$ .
The sum is not computed. The proof produces the limit as $- T$ , where $T$ is the sum of a series that the argument only proves convergent. This is a convergence test and nothing more.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$

Statement

Let $(ε_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , that is the unique sequence of reals with $ε_{0} = 1$ and $ε_{k + 1} = - ε_{k}$ , which is what is usually written $ε_{k} = (- 1)^{k}$ ; let $e$ and $o$ be its even and odd index maps, so that $ε_{e_{j}} = 1$ , $ε_{o_{j}} = - 1$ , and every natural number is $e_{j}$ for exactly one $j$ or $o_{j}$ for exactly one $j$ .

Let $(b_{k})$ be a sequence of reals that is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals); then $b_{k} \geq 0$ for every $k$ . Write $t_{n} : = \sum_{k < n} ε_{k} b_{k}$ for the partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

the series $\sum ε_{k} b_{k}$ converges; write $L$ for its sum;
$t_{e_{j}} \leq L \leq t_{o_{j}}$ for every $j \in N$ , and for every $n \in N$ the sum $L$ lies between the two consecutive partial sums $t_{n}$ and $t_{n + 1}$ ;
$∣ L - t_{n} ∣ \leq b_{n}$ for every $n \in N$ .

Claim 3 is the error bound: the partial sum $t_{n}$ , which uses the $n$ terms $ε_{0} b_{0}, \dots, ε_{n - 1} b_{n - 1}$ , differs from the sum by at most the first term omitted.

Only claim 1 is a corollary of Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges. Claims 2 and 3 are not: they come from the interlacing of the even-index and odd-index partial sums, and that argument is carried out below rather than smuggled into the Dirichlet estimate, which produces no bracketing at all.

Facts & Assumptions

Given: A nonincreasing sequence $(b_{k})$ of reals with $b_{k} \to 0$ , the alternating sequence $(ε_{k})$ with its index maps $e$ and $o$ , and the partial sums $t_{n} = \sum_{k < n} ε_{k} b_{k}$ .

[L1]

The alternating sequence and its index maps: $ε_{0} = 1$ , $ε_{k + 1} = - ε_{k}$ , $∣ ε_{k} ∣ = 1$ ; $e_{0} = 0$ and $e_{j + 1} = e_{j} + 2$ ; $o_{0} = 1$ and $o_{j + 1} = o_{j} + 2$ ; both $e$ and $o$ are strictly increasing; $N$ is the disjoint union of their ranges; $ε_{e_{j}} = 1$ and $ε_{o_{j}} = - 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Nonincreasing means $b_{j} \geq b_{k}$ whenever $j \leq k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L4]

Dirichlet's test: if the partial sums of $\sum x_{k}$ are bounded and $(y_{k})$ is nonincreasing with $y_{k} \to 0$ , then $\sum x_{k} y_{k}$ converges (Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges).

[L5]

A subsequence of a convergent sequence converges to the same limit (Subsequences inherit the limit).

[L6]

Partial sums satisfy $t_{0} = 0$ and $t_{n + 1} = t_{n} + ε_{n} b_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

The principle of induction on $N$ (The principle of mathematical induction).

[L8]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

Proof

technique · direct

1.1

givenL2L3

1.2

Writing $A_{n} = \sum_{k < n} ε_{k}$ , an induction gives that for every $n$ either $A_{n} = 0$ and $ε_{n} = 1$ , or $A_{n} = 1$ and $ε_{n} = - 1$ : at $n = 0$ we have $A_{0} = 0$ and $ε_{0} = 1$ ; and if $A_{n} = 0$ and $ε_{n} = 1$ then $A_{n + 1} = 1$ and $ε_{n + 1} = - 1$ , while if $A_{n} = 1$ and $ε_{n} = - 1$ then $A_{n + 1} = 0$ and $ε_{n + 1} = 1$ . In particular $∣ A_{n} ∣ \leq 1$ for every $n$ .

L1L6L7

1.3

For every $j$ one has $o_{j} = e_{j} + 1$ and $e_{j + 1} = o_{j} + 1$ , by induction: $o_{0} = 1 = e_{0} + 1$ ; and if $o_{j} = e_{j} + 1$ then $e_{j + 1} = e_{j} + 2 = o_{j} + 1$ and $o_{j + 1} = o_{j} + 2 = e_{j + 1} + 1$ .

L1L7

1.4

By [L6], $t_{n + 1} - t_{n} = ε_{n} b_{n}$ for every $n$ ; hence $t_{e_{j} + 1} = t_{e_{j}} + b_{e_{j}}$ and $t_{o_{j} + 1} = t_{o_{j}} - b_{o_{j}}$ .

L1L6

2.1

The partial sums of $\sum ε_{k}$ are bounded by step 1.2 and $(b_{k})$ is nonincreasing with limit $0$ , so $\sum ε_{k} b_{k}$ converges by Dirichlet's test; write $L$ for its sum, so that $t_{n} \to L$ .

step 1.2givenL4

2.2

Using step 1.3, $t_{o_{j}} = t_{e_{j} + 1} = t_{e_{j}} + b_{e_{j}}$ and $t_{e_{j + 1}} = t_{o_{j} + 1} = t_{o_{j}} - b_{o_{j}}$ , so $t_{e_{j + 1}} = t_{e_{j}} + b_{e_{j}} - b_{o_{j}}$ and $t_{o_{j + 1}} = t_{e_{j + 1}} + b_{e_{j + 1}} = t_{o_{j}} - b_{o_{j}} + b_{e_{j + 1}}$ .

step 1.3step 1.4

3.1

Since $e_{j} < o_{j} < e_{j + 1}$ and $(b_{k})$ is nonincreasing, $b_{e_{j}} - b_{o_{j}} \geq 0$ and $b_{e_{j + 1}} - b_{o_{j}} \leq 0$ ; so by step 2.2 the sequence $(t_{e_{j}})_{j}$ is nondecreasing and the sequence $(t_{o_{j}})_{j}$ is nonincreasing.

step 1.3step 2.2L2

3.2

The maps $e$ and $o$ are strictly increasing, so $(t_{e_{j}})_{j}$ and $(t_{o_{j}})_{j}$ are subsequences of $(t_{n})$ and both converge to $L$ .

step 2.1L1L5

4.1

Fix $j$ . For every $m \geq j$ one has $t_{e_{j}} \leq t_{e_{m}}$ , and $(t_{e_{m}})_{m}$ converges to $L$ , so $t_{e_{j}} \leq L$ ; symmetrically $t_{o_{j}} \geq L$ . This is the first half of claim 2.

step 3.1step 3.2L3

5.1

Let $n \in N$ . If $n = e_{j}$ then $t_{n} = t_{e_{j}} \leq L$ and $t_{n + 1} = t_{e_{j} + 1} = t_{o_{j}} \geq L$ ; if $n = o_{j}$ then $t_{n} = t_{o_{j}} \geq L$ and $t_{n + 1} = t_{o_{j} + 1} = t_{e_{j + 1}} \leq L$ . Since every $n$ is of exactly one of these two forms, $L$ always lies between $t_{n}$ and $t_{n + 1}$ , which is the second half of claim 2.

step 1.3step 4.1L1

6.1

Consequently $∣ L - t_{n} ∣ \leq ∣ t_{n + 1} - t_{n} ∣ = ∣ ε_{n} b_{n} ∣ = ∣ ε_{n} ∣ b_{n} = b_{n}$ for every $n$ , using $b_{n} \geq 0$ and $∣ ε_{n} ∣ = 1$ ; this is claim 3.

step 5.1step 1.4step 1.1L1L8∎

Remarks

The two hypotheses are not interchangeable with " $b_{k} \to 0$ " alone. A null sequence that is not monotone can make $\sum ε_{k} b_{k}$ diverge, and the bracketing of step 3.1 is exactly where monotonicity enters; the error bound is false without it. The test as stated is the classical Leibniz criterion.
Why the index maps rather than " $2 j$ " and " $2 j + 1$ ". The even and odd index maps come from The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ together with the parity object itself, and step 1.3 is the only arithmetic needed about them. Rebuilding $(- 1)^{k}$ by a fresh recursion inside this proof, and then proving afresh that the even indices and the odd indices partition $N$ , is precisely what that lemma exists to prevent.
What the test does not give. It produces the sum $L$ as a limit and bounds the error, and it identifies $L$ with no closed expression. For the alternating harmonic series the value is not available at this point in the reading order; see Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals. If $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $(b_{k})$ is monotone (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and bounded, then $\sum a_{k} b_{k}$ converges, and its sum is

$\sum_{k = 0}^{\infty} a_{k} b_{k} = (\sum_{k = 0}^{\infty} a_{k} (b_{k} - b)) + b \sum_{k = 0}^{\infty} a_{k}, b : = \lim_{k} b_{k},$

the limit $b$ existing because a monotone bounded sequence converges (A monotone sequence converges if and only if it is bounded).

Compared with Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges the hypotheses trade places: there $\sum a_{k}$ need only have bounded partial sums while $(b_{k})$ must tend to $0$ ; here $\sum a_{k}$ must converge while $(b_{k})$ need only be monotone with some limit. Neither test implies the other.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ of reals with $\sum a_{k}$ convergent and $(b_{k})$ monotone and bounded, and the partial sums $A_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

A monotone sequence of reals converges if and only if it is bounded (A monotone sequence converges if and only if it is bounded).

[L2]

Monotone means nondecreasing or nonincreasing, and these are the only two possibilities (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

A convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L4]

[L5]

Linearity of series: if $\sum x_{k}$ and $\sum y_{k}$ converge then $\sum (x_{k} + y_{k})$ converges to the sum of the sums, and $\sum c x_{k}$ converges to $c$ times the sum (Convergent series add and scale termwise).

[L6]

Algebra of limits: a convergent sequence minus a constant converges to the limit minus that constant, and multiplying a convergent sequence by $- 1$ negates the limit (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

Proof

technique · cases

1.1

Assume $(b_{k})$ is nonincreasing.

assume-case noninc

1.2

Assume instead $(b_{k})$ is nondecreasing.

assume-case nondec

1.3

In either case $(b_{k})$ is monotone and bounded, so it converges; write $b$ for its limit and put $c_{k} : = b_{k} - b$ , a sequence converging to $0$ .

givenL1L6

1.4

The series $\sum a_{k}$ converges, so its partial sums $A_{n}$ form a convergent sequence and are therefore bounded.

givenL3

2.1

In the case where $(b_{k})$ is nonincreasing, $(c_{k})$ is nonincreasing as well, since it differs from $(b_{k})$ by the constant $b$ .

step 1.1step 1.3L2

2.2

In the case where $(b_{k})$ is nondecreasing, $(- c_{k})$ is nonincreasing and converges to $0$ .

step 1.2step 1.3L2L6

3.1

In the nonincreasing case, $(A_{n})$ is bounded and $(c_{k})$ is nonincreasing with limit $0$ , so $\sum a_{k} c_{k}$ converges by Dirichlet's test.

step 1.4step 2.1L4

3.2

In the nondecreasing case, $(A_{n})$ is bounded and $(- c_{k})$ is nonincreasing with limit $0$ , so $\sum a_{k} (- c_{k})$ converges by Dirichlet's test; multiplying by the constant $- 1$ , $\sum a_{k} c_{k}$ converges.

step 1.4step 2.2L4L5

4.1

So in both cases $\sum a_{k} c_{k}$ converges; and $\sum b a_{k}$ converges, being a constant multiple of the convergent $\sum a_{k}$ .

step 3.1step 3.2L5

5.1

Since $a_{k} b_{k} = a_{k} c_{k} + b a_{k}$ for every $k$ , the series $\sum a_{k} b_{k}$ converges, with sum $\sum_{k = 0}^{\infty} a_{k} c_{k} + b \sum_{k = 0}^{\infty} a_{k}$ , which is the displayed formula.

step 1.3step 4.1L5

6.1

A monotone sequence is nonincreasing or nondecreasing and there is no third possibility, so the two cases cover every hypothesis of the theorem.

step 5.1L2cases-exhaustive∎

Remarks

Both monotonicity directions have to be handled, and only one of them is Dirichlet's hypothesis. Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges requires a nonincreasing factor tending to $0$ . For a nondecreasing bounded $(b_{k})$ the shifted sequence $b_{k} - b$ is nondecreasing and nonpositive, so it is $b - b_{k}$ that Dirichlet's test accepts, and the sign is absorbed afterwards by linearity. Dirichlet's test could equally have been stated with "monotone" in place of "nonincreasing", since the two forms are equivalent for a factor tending to $0$ (Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges, remarks); the proof below takes the nonincreasing form as given and does the sign bookkeeping explicitly, which is why both directions appear.
Boundedness of $(b_{k})$ is used twice. Once through A monotone sequence converges if and only if it is bounded to produce the limit $b$ , and then implicitly in the decomposition $b_{k} = (b_{k} - b) + b$ , which would name nothing if the limit did not exist. Monotone and unbounded is one of the two cases the theorem excludes; the other is bounded and not monotone, and it is that one the companion counterexample to Abel's test on the examples page settles, by showing that dropping monotonicity alone already destroys the conclusion.
The hypothesis on $\sum a_{k}$ cannot be weakened to bounded partial sums. With $a_{k} = (- 1)^{k}$ and $b_{k} = 1$ the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is monotone and bounded, yet $\sum a_{k} b_{k} = \sum (- 1)^{k}$ diverges. What Dirichlet's test adds in that situation is the hypothesis $b_{k} \to 0$ , which fails here.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Rearrangement of a series along a bijection of $N$ , and unconditional convergence

Definition

Let $(a_{k})$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $σ : N \to N$ be a bijection (Injection, surjection, bijection).

Rearrangement. The rearrangement of $(a_{k})$ along $σ$ is the composite sequence $k \mapsto a_{σ (k)}$ , again a function $N \to R$ and so again a sequence of reals. The rearrangement of the series $\sum a_{k}$ along $σ$ is the series $\sum a_{σ (k)}$ of that sequence (Series, partial sums, convergence and the sum, divergence, and the tail series).

A rearrangement uses each term of the original sequence exactly once: injectivity of $σ$ says no term is repeated, surjectivity says none is omitted. That is the whole content of the word, and it is why the definition is stated with a bijection rather than with an informal "reordering".

Unconditional convergence. The series $\sum a_{k}$ converges unconditionally when it converges and, for every bijection $σ : N \to N$ , the rearranged series $\sum a_{σ (k)}$ converges with

$\sum_{k = 0}^{\infty} a_{σ (k)} = \sum_{k = 0}^{\infty} a_{k} .$

Remarks

Unconditional convergence implies convergence, by definition and also by instance. The identity map is a bijection of $N$ and the rearrangement along it is the original sequence, so the clause about all bijections already contains the clause about the series itself.
Rearranging twice is rearranging once. If $σ$ and $τ$ are bijections of $N$ then so is $σ \circ τ$ , and the rearrangement of $(a_{σ (k)})$ along $τ$ is $k \mapsto a_{σ (τ (k))}$ , the rearrangement of $(a_{k})$ along $σ \circ τ$ . Likewise the inverse $σ^{- 1}$ is a bijection, and rearranging along it undoes $σ$ . Both facts are used in Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum.
A weaker-looking condition, which turns out to be the same one. One could ask only that every rearrangement converge, without requiring the sums to agree. Over $R$ that is not weaker: For a series of real numbers, unconditional convergence and absolute convergence are the same property identifies both conditions with absolute convergence (Absolutely convergent and conditionally convergent series, and the general starting index), because The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ produces, for a series that converges but not absolutely, both a rearrangement with a different sum and a rearrangement that does not converge at all.
The definition says nothing about which series have the property. That is the subject of the two theorems that follow, and the answer over $R$ is exactly the absolutely convergent series.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum

Statement

Let $(a_{k})$ be a sequence of reals whose series converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), and let $σ : N \to N$ be a bijection (Injection, surjection, bijection). Then:

$\sum ∣ a_{σ (k)} ∣$ converges, with $\sum_{k = 0}^{\infty} ∣ a_{σ (k)} ∣ = \sum_{k = 0}^{\infty} ∣ a_{k} ∣$ ; that is, the rearranged series again converges absolutely;
$\sum a_{σ (k)}$ converges, with $\sum_{k = 0}^{\infty} a_{σ (k)} = \sum_{k = 0}^{\infty} a_{k} .$

Consequently an absolutely convergent series converges unconditionally (Rearrangement of a series along a bijection of $N$ , and unconditional convergence).

The engine of the proof is a single statement about series of nonnegative terms: for those, the sum is the supremum of the partial sums (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum), a quantity that cannot see the order of the terms. The general case is reduced to that one through the positive and negative parts (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ ), which is why no manipulation of signed finite sums over shuffled index sets occurs anywhere below.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $\sum ∣ a_{k} ∣$ convergent, and a bijection $σ : N \to N$ .

[L1]

Finite sums: $\sum_{k < 0} x_{k} = 0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , and a finite sum may be split at any intermediate index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

Monotonicity of finite sums: if $x_{k} \leq y_{k}$ for all $k < n$ then $\sum_{k < n} x_{k} \leq \sum_{k < n} y_{k}$ ; in particular a finite sum of nonnegative terms is nonnegative (Laws of finite sums and finite products).

[L3]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above, and then the sum is the supremum of that range; in particular every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L4]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L5]

The principle of induction on $N$ (The principle of mathematical induction).

[L6]

A bijection is injective and surjective; $f [S]$ and $f^{- 1} [T]$ denote image and preimage (Injection, surjection, bijection).

[L7]

Positive and negative parts: $a_{k}^{+} = (∣ a_{k} ∣ + a_{k}) / 2$ and $a_{k}^{-} = (∣ a_{k} ∣ - a_{k}) / 2$ are nonnegative, $a_{k} = a_{k}^{+} - a_{k}^{-}$ , $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ , and $\sum ∣ a_{k} ∣$ converges if and only if both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ ).

[L8]

Linearity of series (Convergent series add and scale termwise).

[L9]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[L10]

Unconditional convergence means every rearrangement converges to the same sum (Rearrangement of a series along a bijection of $N$ , and unconditional convergence).

Proof

technique · direct

1.1

Finite domination. For every $n \in N$ the following holds: for every sequence $(c_{k})$ of nonnegative reals, every $Q \in N$ and every injective map $τ$ from ${k : k < n}$ into ${k : k < Q}$ , one has $\sum_{k < n} c_{τ (k)} \leq \sum_{k < Q} c_{k}$ . This is proved by induction on $n$ , the sequence, $Q$ and $τ$ being universally quantified inside the induction statement. At $n = 0$ the left side is the empty sum $0$ and the right side is nonnegative. Assume the statement at $n$ , and let $τ$ be injective from ${k : k < n + 1}$ into ${k : k < Q}$ ; put $p : = τ (n)$ , so $p < Q$ , and let $(c_{k}^{'})$ agree with $(c_{k})$ except that $c_{p}^{'} : = 0$ , again a nonnegative sequence. The restriction of $τ$ to ${k : k < n}$ is injective into ${k : k < Q}$ and never takes the value $p$ , so $c_{τ (k)}^{'} = c_{τ (k)}$ for $k < n$ , and the induction hypothesis gives $\sum_{k < n} c_{τ (k)} = \sum_{k < n} c_{τ (k)}^{'} \leq \sum_{k < Q} c_{k}^{'}$ . Splitting the sum $\sum_{k < Q}$ at $p$ and at $p + 1$ shows $\sum_{k < Q} c_{k}^{'} = \sum_{k < Q} c_{k} - c_{p}$ , so adding $c_{p}$ to both sides gives $\sum_{k < n + 1} c_{τ (k)} \leq \sum_{k < Q} c_{k}$ .

L1L2L5L6

1.2

Bounding index. For every injective $ρ : N \to N$ and every $n \in N$ there is $Q \in N$ with $ρ (k) < Q$ for all $k < n$ : at $n = 0$ take $Q = 0$ , and if $Q$ works for $n$ then the greater of $Q$ and $ρ (n) + 1$ works for $n + 1$ , the order on $N$ being total.

L5L6

1.3

Since $σ$ is a bijection, for every $j \in N$ there is exactly one $k$ with $σ (k) = j$ ; write $σ^{- 1} (j)$ for that $k$ . Then $σ^{- 1}$ is a bijection of $N$ , with $σ (σ^{- 1} (j)) = j$ for every $j$ .

L6choose

1.4

By [L7] both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge; write $U$ and $V$ for their sums. Since $a_{k} = a_{k}^{+} - a_{k}^{-}$ , linearity gives $\sum_{k = 0}^{\infty} a_{k} = U - V$ .

givenL7L8

1.5

The positive and negative parts are defined pointwise from the value of the term, so the positive part of $a_{σ (k)}$ is $a_{σ (k)}^{+}$ and its negative part is $a_{σ (k)}^{-}$ ; both are nonnegative sequences in the index $k$ .

2.1

The nonnegative case, one inequality. Let $(c_{k})$ be a sequence of nonnegative reals with $\sum c_{k}$ convergent of sum $M$ , and let $ρ$ be a bijection of $N$ . For each $n$ pick $Q$ as in step 1.2; then $ρ$ restricted to ${k : k < n}$ is injective into ${k : k < Q}$ , so $\sum_{k < n} c_{ρ (k)} \leq \sum_{k < Q} c_{k} \leq M$ by step 1.1 and [L3]. The terms $c_{ρ (k)}$ are nonnegative, so the partial sums of $\sum c_{ρ (k)}$ are bounded above by $M$ ; hence that series converges, and since each partial sum is at most $M$ its sum is at most $M$ .

step 1.1step 1.2L2L3L4

3.1

The nonnegative case, equality. With $(c_{k})$ , $M$ and $ρ$ as in step 2.1, write $M^{'}$ for the sum of $\sum c_{ρ (k)}$ , so $M^{'} \leq M$ . The sequence $(c_{ρ (k)})_{k}$ is nonnegative with convergent series of sum $M^{'}$ , and its rearrangement along the bijection $ρ^{- 1}$ is $j \mapsto c_{ρ (ρ^{- 1} (j))} = c_{j}$ ; so step 2.1, applied to that sequence and that bijection, gives $M \leq M^{'}$ . Hence $M^{'} = M$ .

step 1.3step 2.1

4.1

Applying step 3.1 to the nonnegative sequence $(∣ a_{k} ∣)$ , whose series converges by hypothesis, and to $σ$ : the series $\sum ∣ a_{σ (k)} ∣$ converges with the same sum as $\sum ∣ a_{k} ∣$ , which is claim 1.

givenstep 3.1

4.2

Applying step 3.1 to $(a_{k}^{+})$ and to $(a_{k}^{-})$ , each with the bijection $σ$ : the series $\sum a_{σ (k)}^{+}$ and $\sum a_{σ (k)}^{-}$ converge, with sums $U$ and $V$ respectively.

step 3.1step 1.4step 1.5

5.1

Since $a_{σ (k)} = a_{σ (k)}^{+} - a_{σ (k)}^{-}$ for every $k$ , linearity gives that $\sum a_{σ (k)}$ converges with sum $U - V$ , which by step 1.4 equals $\sum_{k = 0}^{\infty} a_{k}$ ; this is claim 2.

step 1.4step 1.5step 4.2L8

6.1

The same conclusion is available from claim 1 alone: $\sum ∣ a_{σ (k)} ∣$ converges, so $\sum a_{σ (k)}$ converges; step 5.1 is what identifies its sum.

step 4.1L9

7.1

Claims 1 and 2 hold for an arbitrary bijection $σ$ , so $\sum a_{k}$ converges and every rearrangement of it converges to the same sum, that is, $\sum a_{k}$ converges unconditionally.

step 4.1step 5.1L9L10∎

Remarks

Why the nonnegative case is the whole theorem. For nonnegative terms the sum is the supremum of the set of partial sums (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum), and step 2.1 shows that each partial sum of a rearrangement is bounded by the original sum, and conversely. No cancellation can occur, so nothing depends on the order. Everything genuinely signed in the theorem is handled by Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ , which splits the series into two nonnegative ones.
What step 1.1 is, and why it is proved rather than assumed. It says that a finite sum of nonnegative terms taken along an injective list of indices is at most the sum over an initial segment containing all those indices. This is the one piece of finite combinatorics the theorem needs, and it is not among the laws of Laws of finite sums and finite products, all of which compare sums term by term over the same index range. The proof zeroes out one term at a time, which is what keeps it inside those laws.
The hypothesis cannot be weakened. The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ shows that for a conditionally convergent series every real number, and $\pm \infty$ besides, is the sum of some rearrangement; and For a series of real numbers, unconditional convergence and absolute convergence are the same property turns the two theorems together into an exact characterisation.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$

Statement

Let $(a_{k})$ be a sequence of reals whose series converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index). Let $α, β \in \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined) with $α \leq β$ . Then there is a bijection $σ : N \to N$ (Injection, surjection, bijection) such that the partial sums $T_{n} = \sum_{k < n} a_{σ (k)}$ of the rearranged series (Rearrangement of a series along a bijection of $N$ , and unconditional convergence) satisfy

${lim inf}_{n} T_{n} = α, {lim sup}_{n} T_{n} = β$

(Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ). In particular:

for every $c \in R$ , taking $α = β = c$ , there is a rearrangement of $\sum a_{k}$ that converges with sum $c$ ;
taking $α = β = + \infty$ , there is a rearrangement whose partial sums diverge to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ), and taking $α = β = - \infty$ , one whose partial sums diverge to $- \infty$ ;
taking $α < β$ , there is a rearrangement whose partial sums oscillate, with limit inferior exactly $α$ and limit superior exactly $β$ .

So the sum of a conditionally convergent series is an artefact of the order in which its terms are written, and every prescribed asymptotic behaviour is attainable. Contrast Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, where absolute convergence makes the sum independent of the order.

The construction. Write $P : = {k : a_{k} \geq 0}$ and $N : = {k : a_{k} < 0}$ , which partition $N$ , and enumerate each increasingly as $(p_{i})$ and $(q_{l})$ . Fix real sequences $(u_{j})$ and $(v_{j})$ with $u_{j} \leq v_{j}$ and $u_{j} \leq v_{j + 1}$ for every $j$ ; these are the targets. The rearrangement is produced one index at a time by a greedy rule: while the running sum is at most the current upper target, take the next unused nonnegative term; once it exceeds that target, take negative terms until the running sum falls below the current lower target; then move to the next pair of targets and repeat. Both supplies are inexhaustible, because for a conditionally convergent series both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ diverge to $+ \infty$ (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ ); and the overshoot at each turning point is at most the term just used, which tends to $0$ because $a_{k} \to 0$ (If a series converges then its terms tend to $0$ ). Those two facts are the whole theorem.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $\sum a_{k}$ convergent and $\sum ∣ a_{k} ∣$ divergent; the positive and negative parts $a_{k}^{+}$ , $a_{k}^{-}$ ; the sets $P = {k : a_{k} \geq 0}$ and $N = {k : a_{k} < 0}$ ; and extended reals $α \leq β$ .

[A1]

$P$ and $N$ are disjoint with union $N$ , since the order on $R$ is total; $a_{k}^{+} = a_{k}$ and $a_{k}^{-} = 0$ for $k \in P$ , while $a_{k}^{+} = 0$ and $a_{k}^{-} = - a_{k}$ for $k \in N$ (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ ).

[L1]

For a conditionally convergent series, the partial sums of $\sum a_{k}^{+}$ and of $\sum a_{k}^{-}$ both diverge to $+ \infty$ (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ , Divergence to $+ \infty$ and to $- \infty$ ).

[L2]

The terms of a convergent series tend to $0$ (If a series converges then its terms tend to $0$ ).

[L3]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L4]

The recursion theorem: for a set $A$ , an element $a \in A$ and a function $f : A \to A$ there is a unique $g : N \to A$ with $g (0) = a$ and $g (n + 1) = f (g (n))$ (The recursion theorem).

[L5]

The principle of induction on $N$ (The principle of mathematical induction).

[L6]

Finite sums: $\sum_{k < 0} x_{k} = 0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , splitting at an intermediate index, and $\sum_{k < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

Partial sums of a series and their recursion $s_{n + 1} = s_{n} + a_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L8]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L9]

A bijection is an injective surjection (Injection, surjection, bijection).

[L10]

${lim sup}_{n} x_{n} = \inf {\sup {x_{m} : m \geq n} : n \in N}$ and ${lim inf}_{n} x_{n} = \sup {\inf {x_{m} : m \geq n} : n \in N}$ , both taken in $\overline{R}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L11]

A sequence converges to a real $L$ exactly when its limit inferior and limit superior both equal $L$ , and diverges to $\pm \infty$ exactly when both equal $\pm \infty$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ , Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ ).

[L12]

For nonnegative terms, a series diverges exactly when the range of its partial sums is unbounded above, and then those partial sums diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

Proof

technique · constructive

1.1

Since $\sum a_{k}$ converges, $a_{k} \to 0$ .

givenL2

1.2

For every $K \in N$ there is $k \geq K$ with $k \in P$ : otherwise $a_{k} < 0$ for every $k \geq K$ , so $a_{k}^{+} = 0$ for every $k \geq K$ , so the partial sums of $\sum a_{k}^{+}$ are constant from $K$ on and hence bounded, contradicting [L1]. The same argument with $a_{k}^{-}$ shows that for every $K$ there is $k \geq K$ with $k \in N$ .

A1L1L6L12

2.1

In particular $P$ and $N$ are nonempty, and for every $k$ the sets ${m \in P : m > k}$ and ${m \in N : m > k}$ are nonempty; so by [L3] each has a least element.

step 1.2L3

3.1

Define $p : N \to N$ by $p_{0} : = \min P$ and $p_{i + 1} : = \min {m \in P : m > p_{i}}$ , and $q : N \to N$ by $q_{0} : = \min N$ and $q_{l + 1} : = \min {m \in N : m > q_{l}}$ ; both are legitimate applications of the recursion theorem, the "next element" operations being total functions $N \to N$ by step 2.1. Both $p$ and $q$ take values in $P$ , respectively $N$ , and are strictly increasing.

step 2.1L3L4construct

4.1

An induction gives $p_{i} \geq i$ and $q_{l} \geq l$ for every index, since $p_{0} \geq 0$ and $p_{i + 1} > p_{i} \geq i$ forces $p_{i + 1} \geq i + 1$ .

step 3.1L5

4.2

An induction on $i$ gives $P \cap {k : k < p_{i}} = {p_{i^{'}} : i^{'} < i}$ : at $i = 0$ both sides are empty because $p_{0}$ is the least element of $P$ ; and passing from $i$ to $i + 1$ adds exactly $p_{i}$ , since $p_{i + 1}$ is the least element of $P$ strictly greater than $p_{i}$ , so no element of $P$ lies strictly between them. The same holds for $q$ and $N$ .

step 3.1L5

4.3

Fix real sequences $(u_{j})$ and $(v_{j})$ with $u_{j} \leq v_{j}$ and $u_{j} \leq v_{j + 1}$ for every $j$ . Put $A : = N \times N \times N \times R \times {0, 1}$ , whose elements are written $(i, l, j, s, m)$ , and define $o u t : A \to N$ and $f : A \to A$ by: if $m = 0$ and $s \leq v_{j}$ , then $o u t : = p_{i}$ and $f : = (i + 1, l, j, s + a_{p_{i}}, 0)$ ; if $m = 0$ and $s > v_{j}$ , then $o u t : = q_{l}$ and $f : = (i, l + 1, j, s + a_{q_{l}}, 1)$ ; if $m = 1$ and $s \geq u_{j}$ , then $o u t : = q_{l}$ and $f : = (i, l + 1, j, s + a_{q_{l}}, 1)$ ; if $m = 1$ and $s < u_{j}$ , then $o u t : = p_{i}$ and $f : = (i + 1, l, j + 1, s + a_{p_{i}}, 0)$ . The four cases are exhaustive and mutually exclusive, the order on $R$ being total, so $f$ and $o u t$ are functions.

step 3.1construct

5.1

Every element of $P$ is some $p_{i}$ , and every element of $N$ is some $q_{l}$ : given $k \in P$ , the set ${i : p_{i} > k}$ is nonempty by step 4.1, so it has a least element $i_{0}$ ; $i_{0} \neq 0$ since $p_{0} = \min P \leq k$ , and $p_{i_{0} - 1} \leq k < p_{i_{0}}$ , so $k \in P \cap {m : m < p_{i_{0}}} = {p_{i^{'}} : i^{'} < i_{0}}$ by step 4.2. Together with step 3.1 this says that $p$ is a bijection onto $P$ and $q$ a bijection onto $N$ ; both are injective because they are strictly increasing.

step 3.1step 4.1step 4.2L3L9

5.2

An induction on $i$ gives $\sum_{i^{'} < i} a_{p_{i^{'}}} = \sum_{k < p_{i}} a_{k}^{+}$ : at $i = 0$ every $k < p_{0}$ lies in $N$ , so $a_{k}^{+} = 0$ and both sides are $0$ ; and splitting $\sum_{k < p_{i + 1}} a_{k}^{+}$ at $p_{i}$ and at $p_{i} + 1$ isolates the single term $a_{p_{i}}^{+} = a_{p_{i}}$ , all remaining indices $k$ with $p_{i} < k < p_{i + 1}$ lying in $N$ by step 4.2 and contributing $0$ . The same argument gives $\sum_{l^{'} < l} a_{q_{l^{'}}} = - \sum_{k < q_{l}} a_{k}^{-}$ .

A1step 3.1step 4.2L5L6

5.3

By the recursion theorem let $g : N \to A$ satisfy $g (0) = (0, 0, 0, 0, 0)$ and $g (n + 1) = f (g (n))$ , write $g (n) = (i_{n}, l_{n}, j_{n}, s_{n}, m_{n})$ , and define $σ (n) : = o u t (g (n))$ .

step 4.3L4construct

5.4

For general $α \leq β$ choose real sequences with $u_{j} \leq v_{j}$ and $u_{j} \leq v_{j + 1}$ as follows: if $α, β$ are real, $u_{j} : = α$ and $v_{j} : = β$ ; if $α = - \infty$ and $β$ is real, $u_{j} : = β - (j + 1)$ and $v_{j} : = β$ ; if $α$ is real and $β = + \infty$ , $u_{j} : = α$ and $v_{j} : = α + (j + 1)$ ; if $α = β = + \infty$ , $u_{j} : = j$ and $v_{j} : = j + 1$ ; if $α = β = - \infty$ , $u_{j} : = - (j + 2)$ and $v_{j} : = - (j + 1)$ ; and if $α = - \infty$ , $β = + \infty$ , $u_{j} : = - (j + 1)$ and $v_{j} : = j + 1$ . In every case $(u_{j})$ tends to $α$ and $(v_{j})$ to $β$ in $\overline{R}$ , and both conditions of step 4.3 hold.

step 4.3L11choose

6.1

Hence $\sum_{i^{'} < i} a_{p_{i^{'}}} \to + \infty$ as $i \to \infty$ and $\sum_{l^{'} < l} a_{q_{l^{'}}} \to - \infty$ as $l \to \infty$ : the left-hand sides are the values of the partial sums of $\sum a_{k}^{+}$ , respectively of $- \sum a_{k}^{-}$ , at the strictly increasing indices $p_{i}$ , respectively $q_{l}$ , and by step 4.1 those indices are at least $i$ , respectively $l$ .

step 4.1step 5.2L1

6.2

An induction on $n$ gives $i_{n} + l_{n} = n$ and $s_{n} = \sum_{k < n} a_{σ (k)}$ : both hold at $n = 0$ , and each transition increases exactly one of $i, l$ by one and adds to $s$ exactly the term $a_{σ (n)}$ indexed by the emitted natural. So $s_{n} = T_{n}$ , the $n$ -th partial sum of the rearranged series.

step 4.3step 5.3L5L7

7.1

Consequently, for every $i_{0} \in N$ and every real $M$ there is $i > i_{0}$ with $\sum_{i^{'} = i_{0}}^{i - 1} a_{p_{i^{'}}} > M$ , and for every $l_{0}$ and every real $M$ there is $l > l_{0}$ with $\sum_{l^{'} = l_{0}}^{l - 1} a_{q_{l^{'}}} < M$ ; this is step 6.1 together with splitting of finite sums, the omitted initial block being a fixed real.

step 6.1L6

7.2

An induction on $n$ gives that $σ (n) = p_{i_{n}}$ at every step that increments $i$ , and $σ (n) = q_{l_{n}}$ at every step that increments $l$ ; since $(i_{n})$ and $(l_{n})$ are nondecreasing and increase by one exactly at those steps, distinct steps of the first kind carry distinct values of $i_{n}$ and distinct steps of the second kind distinct values of $l_{n}$ . As $p$ and $q$ are injective with disjoint ranges $P$ and $N$ , the map $σ$ is injective.

step 4.3step 6.2step 5.1L5

8.1

There are infinitely many steps of each kind: if from some step $n_{0}$ on no step increments $l$ , then $m_{n}$ is eventually constantly $0$ , because a step with $m = 1$ that does not increment $l$ sets $m$ to $0$ and a step with $m = 0$ that does not increment $l$ leaves $m$ at $0$ ; then $j_{n}$ is eventually constant, say $j$ , and every subsequent step satisfies $s_{n} \leq v_{j}$ , while by step 7.1 the values $s_{n}$ , which from $n_{0}$ on increase by the successive terms $a_{p_{i}}$ , exceed $v_{j}$ for some $n$ . Symmetrically, if from some step on no step increments $i$ , then $m_{n}$ is eventually constantly $1$ , $j_{n}$ is eventually constant $j$ , every subsequent step satisfies $s_{n} \geq u_{j}$ , and step 7.1 makes $s_{n}$ fall below $u_{j}$ .

step 7.1step 4.3step 6.2L5

9.1

Hence $i_{n} \to \infty$ and $l_{n} \to \infty$ , so every $p_{i}$ and every $q_{l}$ occurs as some $σ (n)$ ; since $P \cup N = N$ and $p, q$ enumerate $P$ and $N$ , the map $σ$ is surjective, and with step 7.2 it is a bijection of $N$ .

A1step 5.1step 7.2step 8.1L9

9.2

Likewise $j_{n} \to \infty$ : if $j_{n}$ were eventually constant $j$ , then from some step on no round is completed, so no step has $m = 1$ and $s < u_{j}$ ; by the argument of step 8.1 the mode is then eventually constant, and either it is $0$ forever, whence $s_{n} \leq v_{j}$ always while $s_{n}$ increases past $v_{j}$ , or it is $1$ forever, whence $s_{n} \geq u_{j}$ always while $s_{n}$ falls below $u_{j}$ .

step 7.1step 4.3step 8.1

10.1

For each $j \geq 1$ let $β_{j}$ be the step at which the mode of round $j$ changes from $0$ to $1$ , that is the unique $n$ with $j_{n} = j$ , $m_{n} = 0$ and $s_{n} > v_{j}$ , and let $α_{j}$ be the step at which round $j$ is completed, the unique $n$ with $j_{n} = j$ , $m_{n} = 1$ and $s_{n} < u_{j}$ ; both exist by step 8.1 and step 9.2, and $α_{j - 1} < β_{j} < α_{j}$ .

step 4.3step 8.1step 9.2choose

11.1

The step $β_{j}$ is preceded, within round $j$ , either by a step that added a term $a_{p_{i}} \geq 0$ to a value $s \leq v_{j}$ , or by the completing step $α_{j - 1}$ of the previous round, which added a term $a_{p_{i}} \geq 0$ to a value $s < u_{j - 1} \leq v_{j}$ . In both situations $v_{j} < T_{β_{j}} \leq v_{j} + a_{p_{i}}$ for the index $i$ used at the immediately preceding step.

step 4.3step 10.1

11.2

Likewise the step $α_{j}$ is preceded within round $j$ by a step that added a term $a_{q_{l}} < 0$ to a value $s \geq u_{j}$ , that step being either an earlier descent step or the switch $β_{j}$ itself, at which $s > v_{j} \geq u_{j}$ ; so $u_{j} - ∣ a_{q_{l}} ∣ \leq T_{α_{j}} < u_{j}$ for the index $l$ used at that step.

step 4.3step 10.1

11.3

For $α_{j - 1} \leq n \leq β_{j}$ the partial sums increase, every step of the climb adding a term $a_{p_{i}} \geq 0$ ; for $β_{j} \leq n \leq α_{j}$ they decrease, every step of the descent adding a term $a_{q_{l}} < 0$ . Hence for every $n$ with $α_{j - 1} \leq n \leq α_{j}$ one has $\min {T_{α_{j - 1}}, T_{α_{j}}} \leq T_{n} \leq T_{β_{j}}$ .

A1step 4.3step 10.1

12.1

Put $δ_{j} : = \max {a_{p_{i (j)}}, ∣ a_{q_{l (j)}} ∣}$ for the two indices appearing in step 11.1 and step 11.2. As $j \to \infty$ those indices tend to infinity, by step 8.1 and step 9.2, so $p_{i (j)} \to \infty$ and $q_{l (j)} \to \infty$ by step 4.1, and $δ_{j} \to 0$ by step 1.1. Thus $v_{j} < T_{β_{j}} \leq v_{j} + δ_{j}$ and $u_{j} - δ_{j} \leq T_{α_{j}} < u_{j}$ for every $j \geq 1$ .

step 1.1step 4.1step 8.1step 9.2step 11.1step 11.2

12.2

Fix $n$ and let $J$ be least with $α_{J - 1} \geq n$ , which exists because the $α_{j}$ are strictly increasing. By step 11.3 every $m \geq α_{J - 1}$ satisfies $T_{m} \leq \sup {T_{β_{j}} : j \geq J}$ , and only the finitely many indices $m$ with $n \leq m < α_{J - 1}$ are unaccounted for; each of those lies in a round of index at most $J - 1$ and so is at most $\max {T_{β_{j}} : 1 \leq j \leq J - 1}$ together with $T_{n}$ itself. Hence $\sup {T_{m} : m \geq n}$ is finite or $+ \infty$ according as $\sup {T_{β_{j}} : j \geq J}$ is, and taking the infimum over $n$ , which drives $J$ to infinity, gives ${lim sup}_{n} T_{n} = {lim sup}_{j} T_{β_{j}}$ .

step 10.1step 11.3L10

13.1

Take $u_{j} = v_{j} = c$ for all $j$ , which satisfies the two conditions of step 4.3. Then $c < T_{β_{j}} \leq c + δ_{j}$ and $c - δ_{j} \leq T_{α_{j}} < c$ , so by step 11.3 every $n$ with $α_{j - 1} \leq n \leq α_{j}$ has $∣ T_{n} - c ∣ \leq \max {δ_{j - 1}, δ_{j}}$ . Given a real $ε > 0$ , choose $J \geq 2$ with $δ_{j} < ε$ for all $j \geq J - 1$ ; then $∣ T_{n} - c ∣ < ε$ for all $n \geq α_{J - 1}$ , so $T_{n} \to c$ and the rearranged series converges with sum $c$ . This is claim 1.

step 12.1step 11.3L8

13.2

Take $v_{j} = j + 1$ and $u_{j} = j$ , which satisfy the two conditions. Then $T_{α_{j}} \geq u_{j} - δ_{j} = j - δ_{j}$ , so by step 11.3 every $n$ with $α_{j - 1} \leq n \leq α_{j}$ has $T_{n} \geq \min {j - 1 - δ_{j - 1}, j - δ_{j}}$ , a quantity that exceeds any prescribed real for all large $j$ ; hence $T_{n} \to + \infty$ . Taking instead $v_{j} = - (j + 1)$ and $u_{j} = - (j + 2)$ , which also satisfy the two conditions, gives $T_{n} \leq T_{β_{j}} \leq v_{j} + δ_{j} = - (j + 1) + δ_{j}$ on the same ranges, hence $T_{n} \to - \infty$ . This is claim 2.

step 12.1step 11.3L8

13.3

By step 12.1 the subsequence $(T_{β_{j}})_{j \geq 1}$ tends to $β$ and $(T_{α_{j}})_{j \geq 1}$ tends to $α$ , in $\overline{R}$ : when the target sequence is real-valued and convergent the two-sided bound of step 12.1 with $δ_{j} \to 0$ gives it, and when the target sequence diverges the one-sided bound does.

step 12.1step 5.4L8L11

14.1

By step 13.3 and [L11], ${lim sup}_{j} T_{β_{j}} = β$ ; so ${lim sup}_{n} T_{n} = β$ . The same argument applied to infima, with $α_{j}$ in place of $β_{j}$ and the lower bound of step 11.3 in place of the upper one, gives ${lim inf}_{n} T_{n} = {lim inf}_{j} T_{α_{j}} = α$ .

step 13.3step 12.2L10L11

15.1

The bijection $σ$ of step 5.3, built from the targets chosen in step 5.4, is therefore a rearrangement of $\sum a_{k}$ whose partial sums have limit inferior $α$ and limit superior $β$ ; claims 1 and 2 are the special cases computed directly in step 13.1 and step 13.2, and claim 3 is the case $α < β$ .

step 9.1step 13.1step 13.2step 14.1discharge-construct∎

Remarks

Only two properties of the series are used. That both part series diverge to $+ \infty$ (Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$ ), which is what keeps the two supplies inexhaustible, and that $a_{k} \to 0$ (If a series converges then its terms tend to $0$ ), which is what makes the overshoot at each turning point vanish. Both hold for every conditionally convergent series and neither holds for an absolutely convergent one, whose part series both converge.
Where the well-ordering principle is used, and where it is not. It appears in step 2.1 and step 3.1, to define the increasing enumerations of $P$ and $N$ , and in step 5.1. It does not appear in the greedy rule: "take terms until the running sum crosses the target" is implemented as a one-step recursion whose state carries the two counters, the round and the running sum, so no least crossing index is ever selected. No choice principle is used anywhere; every object is determined by the data.
Zero terms are not a special case. They are collected into $P$ , so a run of zeros is consumed during a climb without moving the running sum, and the climb still terminates because the tail sums of $\sum_{i} a_{p_{i}}$ are unbounded. Had $P$ been defined as ${k : a_{k} > 0}$ , the zero-indexed terms would have had to be inserted separately for $σ$ to be surjective.
The oscillating case is genuinely more than the two divergences. With $α < β$ both finite, the partial sums visit every neighbourhood of $α$ and of $β$ infinitely often and are eventually confined to a neighbourhood of $[α, β]$ ; the subsequential limit set of $(T_{n})$ is then the whole interval, though nothing on this page needs that refinement.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

For a series of real numbers, unconditional convergence and absolute convergence are the same property

Statement

Let $(a_{k})$ be a sequence of reals. The following are equivalent.

$\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).
$\sum a_{k}$ converges unconditionally (Rearrangement of a series along a bijection of $N$ , and unconditional convergence).
$\sum a_{k}$ converges and every rearrangement of it converges, with no requirement that the sums agree.

So over $R$ there is nothing between absolute and conditional convergence: a convergent series either may be reordered freely, sum and all, or else has a rearrangement that fails to converge at all.

This is a statement about $R$ , and nothing here says how much of it survives elsewhere. Whether the equivalence of 1 and 2 holds for series of vectors is a question this library cannot pose at this point in the reading order, since it has no notion of a convergent series of vectors; it is raised, and left open, in The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order. No claim about any space other than $R$ is made or used here.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals.

[L1]

An absolutely convergent series converges unconditionally: every rearrangement converges, to the same sum (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum).

[L2]

If $\sum a_{k}$ converges conditionally then for every $α \leq β$ in the extended reals there is a rearrangement whose partial sums have those as limit inferior and limit superior; in particular there is one whose partial sums diverge to $+ \infty$ (The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ ).

[L3]

Unconditional convergence means: the series converges, and every rearrangement converges to the same sum (Rearrangement of a series along a bijection of $N$ , and unconditional convergence).

[L4]

A series converges absolutely when $\sum ∣ a_{k} ∣$ converges, and conditionally when it converges while $\sum ∣ a_{k} ∣$ does not; a convergent series is exactly one of the two (Absolutely convergent and conditionally convergent series, and the general starting index).

[L5]

A sequence diverging to $+ \infty$ does not converge: if $x_{n} \to + \infty$ and also $x_{n} \to L$ , then eventually $x_{n} > L + 1$ and eventually $∣ x_{n} - L ∣ < 1$ , which are incompatible (Divergence to $+ \infty$ and to $- \infty$ , Limits and Cauchy sequences of reals, Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · direct

1.1

Assume 1. Then by [L1] the series converges and every rearrangement converges to the same sum, which is 2.

L1L3

1.2

Assume 2. Then in particular the series converges and every rearrangement converges, which is 3.

1.3

Assume 3, and suppose $\sum a_{k}$ did not converge absolutely. Since it converges, it would then converge conditionally.

2.1

In that situation [L2] supplies a bijection $σ$ of $N$ for which the partial sums of $\sum a_{σ (k)}$ diverge to $+ \infty$ , and such a series does not converge; this contradicts the assumption that every rearrangement converges.

step 1.3L2L5

3.1

Hence under 3 the series converges absolutely, which is 1.

step 1.3step 2.1L4

4.1

The implications 1 to 2, 2 to 3 and 3 to 1 close the cycle, so the three statements are equivalent.

step 1.1step 1.2step 3.1∎

Remarks

Statement 3 is the reason the corollary is worth recording. It says that merely asking every rearrangement to converge already forces absolute convergence, so the apparently weaker demand is not weaker at all. What makes that work is the strength of The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ : it produces not only rearrangements with prescribed sums but rearrangements with no sum.
Two of the three implications are cheap. The content is in 3 implies 1, and its only ingredient is the Riemann series theorem. The implication 1 implies 2 is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum verbatim, and 2 implies 3 is a weakening.
Where the dividing line sits. By Absolutely convergent and conditionally convergent series, and the general starting index a convergent series is absolutely or conditionally convergent and not both, so the corollary may be read as: the conditionally convergent series are exactly the convergent series that are not unconditionally convergent. The alternating harmonic series is the standard inhabitant of that class; see FALSE: every rearrangement of a convergent series converges, and to the same sum.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum

Statement

Let $(a_{k})$ be a sequence of reals whose series converges (Series, partial sums, convergence and the sum, divergence, and the tail series), with sum $S$ , and let $n : N \to N$ be strictly increasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) with $n_{0} = 0$ . Define the blocks

$B_{j} : = \sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k} (j \in N),$

each a finite sum of $n_{j + 1} - n_{j} \geq 1$ consecutive terms (Finite sums and finite products, by recursion). Then $\sum B_{j}$ converges, with

$\sum_{j = 0}^{\infty} B_{j} = S .$

The proof shows more, and the extra is what makes the theorem trivial once seen: the $m$ -th partial sum of $\sum B_{j}$ is exactly $s_{n_{m}}$ , the $n_{m}$ -th partial sum of $\sum a_{k}$ . Grouping does not produce a new series so much as a subsequence of the old partial sums.

The converse fails. A grouped series may converge while the original diverges, and FALSE: if some grouping of a series converges then the series itself converges records that, with the witness on the companion page. What the theorem needs is convergence of $\sum a_{k}$ as a hypothesis, and $n_{0} = 0$ , without which the first block would omit the terms before $n_{0}$ .

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $\sum a_{k}$ convergent of sum $S$ and partial sums $s_{n} = \sum_{k < n} a_{k}$ ; a strictly increasing $n : N \to N$ with $n_{0} = 0$ ; and the blocks $B_{j} = \sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ .

[L1]

Finite sums: $\sum_{k < 0} x_{k} = 0$ and $\sum_{k < m + 1} x_{k} = \sum_{k < m} x_{k} + x_{m}$ (Finite sums and finite products, by recursion).

[L2]

Splitting: if $m \leq n$ then $\sum_{k < n} a_{k} = \sum_{k < m} a_{k} + \sum_{k = m}^{n - 1} a_{k}$ (Laws of finite sums and finite products).

[L3]

The partial sums of a series and the meaning of its sum (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

A subsequence of a convergent sequence converges to the same limit; a subsequence is indexed by a strictly increasing map $N \to N$ (Subsequences inherit the limit, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L5]

The principle of induction on $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

Since $n$ is strictly increasing, $n_{j} < n_{j + 1}$ for every $j$ , so each block $B_{j}$ is a finite sum over a nonempty range of indices and is a well-determined real.

givenL1L2

1.2

The map $m \mapsto n_{m}$ is strictly increasing, so $(s_{n_{m}})_{m}$ is a subsequence of the convergent sequence $(s_{n})$ and therefore converges to $S$ .

givenL3L4

2.1

An induction on $m$ gives $\sum_{j < m} B_{j} = s_{n_{m}}$ for every $m \in N$ : at $m = 0$ the left side is the empty sum $0$ and the right side is $s_{n_{0}} = s_{0} = 0$ ; and if $\sum_{j < m} B_{j} = s_{n_{m}}$ then $\sum_{j < m + 1} B_{j} = s_{n_{m}} + B_{m} = \sum_{k < n_{m}} a_{k} + \sum_{k = n_{m}}^{n_{m + 1} - 1} a_{k} = \sum_{k < n_{m + 1}} a_{k} = s_{n_{m + 1}}$ , the middle equality being splitting at $n_{m} \leq n_{m + 1}$ .

givenstep 1.1L1L2L5

3.1

By step 2.1 the partial sums of $\sum B_{j}$ are precisely the terms $s_{n_{m}}$ , so $\sum B_{j}$ converges with sum $S$ .

step 2.1step 1.2L3∎

Remarks

Why $n_{0} = 0$ is a hypothesis and not a normalisation. If $n_{0} > 0$ the same computation gives $\sum_{j < m} B_{j} = s_{n_{m}} - s_{n_{0}}$ , so the grouped series converges to $S - s_{n_{0}}$ : the terms $a_{0}, \dots, a_{n_{0} - 1}$ are simply omitted. The theorem as stated is the case where nothing is omitted.
Blocks may be as long as one likes, and the theorem is indifferent. No bound on $n_{j + 1} - n_{j}$ is assumed, and none is needed: the argument never looks inside a block. This is exactly what fails in the converse direction, where the cancellation hidden inside long blocks is what the grouped series cannot see.
The result also gives the associativity one expects of a convergent series. Any two groupings of a convergent series have the same sum, both being $S$ ; so one may insert brackets at will, though never remove them (FALSE: if some grouping of a series converges then the series itself converges).

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$

Definition

Let $(a_{k})$ and $(b_{k})$ be sequences of reals (Series, partial sums, convergence and the sum, divergence, and the tail series). The Cauchy product of $\sum a_{k}$ and $\sum b_{k}$ is the series $\sum c_{n}$ of the sequence

$c_{n} : = \sum_{k = 0}^{n} a_{k} b_{n - k} (n \in N),$

a finite sum of $n + 1$ terms in the sense of Finite sums and finite products, by recursion. Each index $n - k$ occurring here is a natural number, because $k$ runs over $0, \dots, n$ ; and $c_{0} = a_{0} b_{0}$ .

The definition uses only the two sequences of terms. No convergence is assumed and none is asserted: $\sum c_{n}$ is a series formed from $(a_{k})$ and $(b_{k})$ , and whether it converges, and to what, is the subject of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ and If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ , while FALSE: the Cauchy product of two convergent series converges shows that convergence of both factors is not enough.

Why these coefficients. Reading $\sum a_{k} x^{k}$ and $\sum b_{k} x^{k}$ as formal power series and multiplying them term by term, the coefficient of $x^{n}$ collects exactly the products $a_{k} b_{n - k}$ with $k + (n - k) = n$ . So $c_{n}$ is the coefficient one is forced to write down if the product of two series is to behave like the product of two polynomials, and the results on this page say when that formal operation computes the product of the two sums.

Remarks

Only the two sequences of terms enter. The construction is a rule on sequences, and every result below is stated for the sequence $(c_{n})$ it produces. The Cauchy product of $\sum b_{k}$ with $\sum a_{k}$ is formed by the same rule with the roles exchanged, giving $\sum_{k = 0}^{n} b_{k} a_{n - k}$ ; that this is the same number as $c_{n}$ is the reversal invariance of a finite sum, which is not among the laws of Laws of finite sums and finite products and is not used anywhere on this page. Each statement below therefore says which factor carries which hypothesis, rather than appealing to symmetry.
The definition is stated for series indexed from $0$ , as every series on this page is (Series, partial sums, convergence and the sum, divergence, and the tail series). For families from a general starting index the Cauchy product is formed after shifting both families to $N$ , as Series, partial sums, convergence and the sum, divergence, and the tail series prescribes; the shift changes which products appear in $c_{n}$ , so the starting indices have to be said, and they are said wherever this construction is used below.
Nothing here is a product of sums. The symbol $\sum c_{n}$ names a new series built from the terms, not the number $(\sum_{k = 0}^{\infty} a_{k}) (\sum_{k = 0}^{\infty} b_{k})$ , which may not even be defined. Identifying the two is a theorem with hypotheses.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals, let $A_{n} = \sum_{i < n} a_{i}$ and $B_{m} = \sum_{j < m} b_{j}$ be their partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series), and let $(c_{n})$ be their Cauchy product, $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ). Then:

A finite identity, holding for arbitrary sequences. For every $N \in N$ , $\sum_{n < N} c_{n} = \sum_{i < N} a_{i} B_{N - i} .$
Mertens' theorem. If $\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index) and $\sum b_{k}$ converges, then $\sum a_{k}$ converges, say to $A$ , and, writing $B$ for the sum of $\sum b_{k}$ , the Cauchy product $\sum c_{n}$ converges with $\sum_{n = 0}^{\infty} c_{n} = A B .$

Claim 1 carries no hypothesis at all and is used again, for the sequences $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ , in If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ ; that is why it is stated as part of the theorem rather than buried in the proof.

The hypotheses are not symmetric, and that is the point. Only one of the two series is required to converge absolutely; the other need only converge. Requiring convergence of both and nothing more is not enough, as FALSE: the Cauchy product of two convergent series converges shows.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ of reals, their partial sums $A_{n}$ and $B_{m}$ , and their Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[L1]

Finite sums: $\sum_{k < 0} x_{k} = 0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , and $\sum_{k = 0}^{n} x_{k} = \sum_{k < n + 1} x_{k}$ (Finite sums and finite products, by recursion).

[L2]

Finite sums are additive, are scaled by a constant factor, may be split at an intermediate index, and are monotone in their terms (Laws of finite sums and finite products).

[L3]

Partial sums of a series, and the meaning of its sum (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

The principle of induction on $N$ (The principle of mathematical induction).

[L5]

$∣ \sum_{k < n} x_{k} ∣ \leq \sum_{k < n} ∣ x_{k} ∣$ (Triangle inequality for finite sums).

[L6]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[L7]

For a series of nonnegative terms, every partial sum is at most the sum, and the partial sums converge to it (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L8]

A convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L9]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L10]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges; absolute convergence of $\sum a_{k}$ means convergence of $\sum ∣ a_{k} ∣$ (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct

1.1

Claim 1 holds, by induction on $N$ . At $N = 0$ both sides are empty sums, hence $0$ . Assume it at $N$ . By [L1], $\sum_{n < N + 1} c_{n} = \sum_{n < N} c_{n} + c_{N}$ and $c_{N} = \sum_{k < N + 1} a_{k} b_{N - k} = \sum_{k < N} a_{k} b_{N - k} + a_{N} b_{0}$ . On the other side, $\sum_{i < N + 1} a_{i} B_{N + 1 - i} = \sum_{i < N} a_{i} B_{N + 1 - i} + a_{N} B_{1}$ , where $B_{1} = b_{0}$ by [L1] and $B_{N + 1 - i} = B_{N - i} + b_{N - i}$ for every $i \leq N$ , again by [L1]; so additivity gives $\sum_{i < N + 1} a_{i} B_{N + 1 - i} = \sum_{i < N} a_{i} B_{N - i} + \sum_{i < N} a_{i} b_{N - i} + a_{N} b_{0}$ . Substituting the induction hypothesis into the first term and recognising the last two as $c_{N}$ closes the induction.

L1L2L4

1.2

Assume the hypotheses of claim 2. Since $\sum ∣ a_{k} ∣$ converges, $\sum a_{k}$ converges; write $A$ for its sum, so $A_{n} \to A$ , and write $L$ for the sum of $\sum ∣ a_{k} ∣$ , so that $P_{N} : = \sum_{k < N} ∣ a_{k} ∣$ satisfies $P_{N} \leq L$ for every $N$ and $P_{N} \to L$ .

givenL3L7L10

1.3

Write $B$ for the sum of $\sum b_{k}$ and $β_{m} : = B_{m} - B$ , so that $β_{m} \to 0$ ; being convergent, $(β_{m})$ is bounded, and we fix a real $C \geq 1$ with $∣ β_{m} ∣ \leq C$ for every $m$ .

givenL3L8L9choose

2.1

By claim 1 and additivity, for every $N$ , $\sum_{n < N} c_{n} = \sum_{i < N} a_{i} (B + β_{N - i}) = B A_{N} + R_{N}$ , where $R_{N} : = \sum_{i < N} a_{i} β_{N - i}$ .

step 1.1step 1.3L2

2.2

Let $ε > 0$ be real. Since $P_{N} \to L$ , fix $M \in N$ with $L - P_{M} < ε / (2 C)$ ; since $β_{m} \to 0$ , fix $K \in N$ with $∣ β_{m} ∣ < ε / (2 (L + 1))$ for all $m \geq K$ . Both quotients are legitimate, $C \geq 1$ and $L + 1 \geq 1$ being positive.

step 1.2step 1.3choose

3.1

For $N \geq M + K$ , the triangle inequality and splitting at $M$ give $∣ R_{N} ∣ \leq \sum_{i < N} ∣ a_{i} ∣ ∣ β_{N - i} ∣ = \sum_{i < M} ∣ a_{i} ∣ ∣ β_{N - i} ∣ + \sum_{i = M}^{N - 1} ∣ a_{i} ∣ ∣ β_{N - i} ∣$ .

step 2.1L2L5L6

4.1

In the first of those sums $i < M$ and $N \geq M + K$ , so $N - i > N - M \geq K$ and in particular $N - i \geq K$ , whence $∣ β_{N - i} ∣ < ε / (2 (L + 1))$ ; monotonicity of finite sums then bounds it by $ε P_{M} / (2 (L + 1)) \leq ε L / (2 (L + 1)) < ε / 2$ .

step 2.2step 3.1step 1.2L2

4.2

In the second sum every factor $∣ β_{N - i} ∣$ is at most $C$ , so it is bounded by $C \sum_{i = M}^{N - 1} ∣ a_{i} ∣ = C (P_{N} - P_{M}) \leq C (L - P_{M}) < ε / 2$ .

step 2.2step 3.1step 1.2step 1.3L2

5.1

Hence $∣ R_{N} ∣ < ε$ for every $N \geq M + K$ ; as $ε > 0$ was arbitrary, $R_{N} \to 0$ .

step 3.1step 4.1step 4.2L3

6.1

By step 2.1, step 1.2 and step 5.1 the partial sums of $\sum c_{n}$ satisfy $\sum_{n < N} c_{n} = B A_{N} + R_{N} \to B A + 0 = A B$ , so $\sum c_{n}$ converges with sum $A B$ , which is claim 2.

step 1.2step 2.1step 5.1L9∎

Remarks

Where absolute convergence of $\sum a_{k}$ is used. Twice, and both times to control a tail of $\sum ∣ a_{k} ∣$ : in step 2.2, to make the far block of the splitting small uniformly in $N$ , and in step 4.1, where $P_{M} \leq L$ bounds the near block. Mere convergence of $\sum a_{k}$ gives no such control, since the tail of a conditionally convergent series is small only after cancellation, and the factors $β_{N - i}$ destroy the cancellation.
The identity of claim 1 is a rectangle folded into a triangle. It says that summing the products $a_{i} b_{j}$ over the triangle $i + j < N$ by antidiagonals gives the same result as summing them row by row, $\sum_{i < N} a_{i} \sum_{j < N - i} b_{j}$ . The induction proves exactly that, and it needs no hypothesis because both sides are finite sums.
Abel's stronger theorem is not available here. If $\sum a_{k}$ , $\sum b_{k}$ and $\sum c_{n}$ all converge, then the sum of $\sum c_{n}$ is $A B$ without any absolute convergence; but the standard proof runs through power series and Abel's limit theorem, which are later in the reading order. Mertens' theorem is what this page can prove, and its hypotheses are what If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ inherits.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals whose series both converge absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), with sums $A$ and $B$ , and let $(c_{n})$ be their Cauchy product (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ). Then $\sum c_{n}$ converges absolutely, and

$\sum_{n = 0}^{\infty} c_{n} = A B .$

Moreover $\sum_{n = 0}^{\infty} ∣ c_{n} ∣ \leq (\sum_{k = 0}^{\infty} ∣ a_{k} ∣) (\sum_{k = 0}^{\infty} ∣ b_{k} ∣)$ .

Combined with Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum this says that within the absolutely convergent series the product behaves exactly as one would want: it converges, its sum is the product of the sums, and neither factor's order nor the product's order matters.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ with $\sum ∣ a_{k} ∣$ and $\sum ∣ b_{k} ∣$ convergent, sums $L_{a}$ and $L_{b}$ respectively, partial sums $P_{N} = \sum_{k < N} ∣ a_{k} ∣$ and $Q_{m} = \sum_{j < m} ∣ b_{j} ∣$ , and the Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[L1]

The finite identity of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , claim 1: for arbitrary sequences $(x_{k})$ , $(y_{k})$ with partial sums $Y_{m} = \sum_{j < m} y_{j}$ and Cauchy product $(z_{n})$ , one has $\sum_{n < N} z_{n} = \sum_{i < N} x_{i} Y_{N - i}$ for every $N$ .

[L2]

Mertens' theorem, claim 2 of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ : if $\sum x_{k}$ converges absolutely and $\sum y_{k}$ converges, their Cauchy product converges to the product of the sums.

[L3]

$∣ \sum_{k < n} x_{k} ∣ \leq \sum_{k < n} ∣ x_{k} ∣$ (Triangle inequality for finite sums).

[L4]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[L5]

Finite sums are monotone in their terms and scale by a constant factor; the empty sum is $0$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above, and then every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

Direct comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L8]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Proof

technique · direct

1.1

Both $L_{a}$ and $L_{b}$ are nonnegative, and $P_{N} \leq L_{a}$ and $Q_{m} \leq L_{b}$ for all $N$ and $m$ , the terms $∣ a_{k} ∣$ and $∣ b_{j} ∣$ being nonnegative.

givenL4L6

1.2

Put $γ_{n} : = \sum_{k = 0}^{n} ∣ a_{k} ∣ ∣ b_{n - k} ∣$ , the Cauchy product of the sequences $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ ; every $γ_{n}$ is nonnegative.

givenL4L5

2.1

For every $n$ , $∣ c_{n} ∣ = ∣ \sum_{k = 0}^{n} a_{k} b_{n - k} ∣ \leq \sum_{k = 0}^{n} ∣ a_{k} b_{n - k} ∣ = γ_{n}$ .

step 1.2L3L4

2.2

Applying [L1] to $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ gives $\sum_{n < N} γ_{n} = \sum_{i < N} ∣ a_{i} ∣ Q_{N - i}$ for every $N$ .

step 1.2L1

3.1

Since $0 \leq Q_{N - i} \leq L_{b}$ and $∣ a_{i} ∣ \geq 0$ , monotonicity and scaling give $\sum_{i < N} ∣ a_{i} ∣ Q_{N - i} \leq \sum_{i < N} ∣ a_{i} ∣ L_{b} = L_{b} P_{N} \leq L_{b} L_{a}$ for every $N$ .

step 1.1step 2.2L5

4.1

So $\sum γ_{n}$ is a series of nonnegative terms whose partial sums are bounded above by $L_{a} L_{b}$ ; it therefore converges, with sum at most $L_{a} L_{b}$ .

step 1.2step 3.1L6

5.1

By step 2.1 and comparison, $\sum ∣ c_{n} ∣$ converges, and its sum is at most that of $\sum γ_{n}$ , hence at most $L_{a} L_{b}$ ; that is, $\sum c_{n}$ converges absolutely and satisfies the displayed bound.

step 2.1step 4.1L6L7

6.1

The hypotheses of Mertens' theorem hold, $\sum a_{k}$ converging absolutely and $\sum b_{k}$ converging by step 1.1 and [L8]; so $\sum c_{n}$ converges with sum $A B$ .

givenL2L8∎

Remarks

Only claim 1 of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ is reused, and it is reused for a different pair of sequences. The identity there is proved for arbitrary real sequences and carries no convergence hypothesis, which is exactly what allows it to be applied here to $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ . Nothing about the absolute values is reproved.
The bound is not an equality. The sum of $\sum ∣ c_{n} ∣$ can be strictly less than $L_{a} L_{b}$ , since cancellation inside each $c_{n}$ is invisible to $γ_{n}$ ; the inequality is all that is claimed and all that is needed.
Absolute convergence of both factors is a genuine strengthening. Mertens' theorem already gives $\sum c_{n} = A B$ with only one factor absolutely convergent; what the second hypothesis buys is that the product series is itself absolutely convergent, hence unconditionally convergent (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum), so that its terms may be reordered in turn.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value

Statement

Let $a : N \times N \to R$ be a doubly indexed array of reals, written $a_{i j}$ . Assume:

(H) for every $i$ the series $\sum_{j} ∣ a_{i j} ∣$ converges, with sum $A_{i}$ ; and the series $\sum_{i} A_{i}$ converges, with sum $L$ .

Then, with $J : N \to N \times N$ any bijection ( $N \times N \approx N$ , Injection, surjection, bijection):

$\sum_{n} a_{J (n)}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), and its sum $S$ is the same for every such bijection (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum);
for every $i$ the series $\sum_{j} a_{i j}$ converges, say to $R_{i}$ ; the series $\sum_{i} R_{i}$ converges absolutely; and $\sum_{i = 0}^{\infty} R_{i} = S$ ;
for every $j$ the series $\sum_{i} ∣ a_{i j} ∣$ converges and $\sum_{i} a_{i j}$ converges, say to $C_{j}$ ; the series $\sum_{j} C_{j}$ converges absolutely; and $\sum_{j = 0}^{\infty} C_{j} = S$ .

In particular the two iterated sums exist and agree:

$\sum_{i = 0}^{\infty} (\sum_{j = 0}^{\infty} a_{i j}) = \sum_{j = 0}^{\infty} (\sum_{i = 0}^{\infty} a_{i j}) = \sum_{n = 0}^{\infty} a_{J (n)} .$

The hypothesis is on the absolute values, and it is stated as an iterated condition, not as an unqualified "double sum". Each row must be absolutely summable, and the row totals must themselves be summable. Without it the two iterated sums may both exist and differ, which is FALSE: whenever both iterated sums of a double array exist, they are equal.

Facts & Assumptions

Given: An array $a : N \times N \to R$ satisfying (H), with row totals $A_{i}$ and $L = \sum_{i = 0}^{\infty} A_{i}$ , and a bijection $J : N \to N \times N$ .

[L1]

Finite sums: the empty sum is $0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , finite sums are additive, monotone in their terms, and may be split at any intermediate index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above; then the sum is the supremum of that range, every partial sum is at most the sum, and the partial sums converge to it (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L3]

Direct comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L4]

$∣ \sum_{k < n} x_{k} ∣ \leq \sum_{k < n} ∣ x_{k} ∣$ (Triangle inequality for finite sums).

[L5]

Absolute value: $∣ x ∣ \geq 0$ , $- ∣ x ∣ \leq x \leq ∣ x ∣$ , and $∣ x ∣ = 0$ exactly when $x = 0$ (Basic properties of the absolute value).

[L6]

The principle of induction on $N$ (The principle of mathematical induction).

[L7]

A bijection is an injective surjection; $N \times N$ admits a bijection with $N$ (Injection, surjection, bijection, $N \times N \approx N$ ).

[L8]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, Absolutely convergent and conditionally convergent series, and the general starting index).

[L9]

An absolutely convergent series has the same sum along every rearrangement (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Rearrangement of a series along a bijection of $N$ , and unconditional convergence).

[L10]

Algebra of limits, and limits preserve non-strict inequalities holding eventually (Algebra of limits: sums, scalar multiples, products and quotients, Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Rectangles are bounded by $L$ . For all $P, Q \in N$ one has $\sum_{i < P} \sum_{j < Q} ∣ a_{i j} ∣ \leq \sum_{i < P} A_{i} \leq L$ , since each inner sum is a partial sum of the convergent nonnegative series $\sum_{j} ∣ a_{i j} ∣$ and so is at most $A_{i}$ , and finite sums are monotone.

givenL1L2

1.2

Single points. Let $d : N \times N \to R$ vanish except at one pair $(p, q)$ , let $N \in N$ and let $ρ$ be injective on ${n : n < N}$ with values in $N \times N$ . If $(p, q) = ρ (n_{0})$ for some (necessarily unique) $n_{0} < N$ , then $\sum_{n < N} d_{ρ (n)} = d_{p q}$ ; otherwise $\sum_{n < N} d_{ρ (n)} = 0$ . Both follow by splitting the sum at $n_{0}$ and at $n_{0} + 1$ , all remaining terms being $0$ .

L1L7

1.3

List dominated by a rectangle. For every $N \in N$ , every array $(c_{i j})$ of nonnegative reals, all $P, Q \in N$ and every injective $ρ$ on ${n : n < N}$ with values in ${(i, j) : i < P, j < Q}$ , one has $\sum_{n < N} c_{ρ (n)} \leq \sum_{i < P} \sum_{j < Q} c_{i j}$ . Induction on $N$ , everything else universally quantified: at $N = 0$ the left side is $0$ and the right side is nonnegative; and passing from $N$ to $N + 1$ , put $(p, q) : = ρ (N)$ and let $c^{''}$ agree with $c$ except that $c_{p q}^{''} : = 0$ , so that the induction hypothesis applied to $c^{''}$ and $ρ$ restricted gives $\sum_{n < N} c_{ρ (n)} \leq \sum_{i < P} \sum_{j < Q} c_{i j} - c_{p q}$ , the subtraction coming from splitting the outer sum at $p$ and the inner one at $q$ ; adding $c_{p q}$ closes the induction.

L1L6

1.4

Bounding indices. For every $N$ there are $P, Q$ with $J (n) \in {(i, j) : i < P, j < Q}$ for all $n < N$ ; and for all $P, Q$ there is $N$ with ${(i, j) : i < P, j < Q} \subseteq {J (n) : n < N}$ . Both are inductions using that the order on $N$ is total, so that finitely many naturals have a strict upper bound; the second uses surjectivity of $J$ to name, for each pair, the index mapping onto it.

L6L7

1.5

For every $i$ the series $\sum_{j} a_{i j}$ converges, since $\sum_{j} ∣ a_{i j} ∣$ does; write $R_{i}$ for its sum, so $∣ R_{i} ∣ \leq A_{i}$ by [L4] and [L10]. Hence $\sum_{i} ∣ R_{i} ∣$ converges by comparison with $\sum_{i} A_{i}$ , and $\sum_{i} R_{i}$ converges absolutely.

givenL2L3L4L8L10

1.6

Let $ε > 0$ be real. Choose $P_{0} \geq 1$ with $L - \sum_{i < P_{0}} A_{i} < ε$ , possible because the partial sums of $\sum_{i} A_{i}$ converge to $L$ ; then choose, for each $i < P_{0}$ , an index $Q_{i}$ with $A_{i} - \sum_{j < Q_{i}} ∣ a_{i j} ∣ < ε / P_{0}$ , and let $Q_{0}$ be an upper bound of the finitely many $Q_{i}$ , so that $A_{i} - \sum_{j < Q_{0}} ∣ a_{i j} ∣ < ε / P_{0}$ for every $i < P_{0}$ .

givenL2L6choose

2.1

Rectangle to list. Let $c$ be an array, let $P, Q, N \in N$ and let $ρ$ be injective on ${n : n < N}$ with ${(i, j) : i < P, j < Q} \subseteq {ρ (n) : n < N}$ . Let $c^{'}$ agree with $c$ on that rectangle and vanish off it. Then $\sum_{i < P} \sum_{j < Q} c_{i j} = \sum_{n < N} c_{ρ (n)}^{'}$ . This is proved by induction on $P$ , with an inner induction on $Q$ : enlarging the rectangle by one column adds the single term $c_{P Q}$ to the left side, and changes $c^{'}$ by an array vanishing except at $(P, Q)$ , which by step 1.2 adds exactly $c_{P Q}$ to the right side; at $P = 0$ or $Q = 0$ both sides are $0$ .

step 1.2L1L6

2.2

By step 1.3 and step 1.4, every partial sum $\sum_{n < N} ∣ a_{J (n)} ∣$ is at most $\sum_{i < P} \sum_{j < Q} ∣ a_{i j} ∣ \leq L$ ; hence $\sum_{n} ∣ a_{J (n)} ∣$ converges, with sum $Λ \leq L$ , and $\sum_{n} a_{J (n)}$ converges, say to $S$ . Any two bijections $N \to N \times N$ differ by a bijection of $N$ , so by [L9] the value $S$ does not depend on $J$ ; this is claim 1.

step 1.1step 1.3step 1.4L2L8L9

2.3

Write $D : = \sum_{i < P_{0}} \sum_{j < Q_{0}} a_{i j}$ and $E : = \sum_{i < P_{0}} \sum_{j < Q_{0}} ∣ a_{i j} ∣$ . By step 1.6 and monotonicity, $E > \sum_{i < P_{0}} (A_{i} - ε / P_{0}) = \sum_{i < P_{0}} A_{i} - ε > L - 2 ε$ , so $L - E < 2 ε$ .

step 1.6L1

2.4

By step 1.4 fix $N$ with ${(i, j) : i < P_{0}, j < Q_{0}} \subseteq {J (n) : n < N}$ , and by step 1.4 again fix $P \geq P_{0}$ , $Q \geq Q_{0}$ with $J (n)$ in the rectangle ${(i, j) : i < P, j < Q}$ for all $n < N$ .

step 1.4choose

2.5

The transposed array $a_{i j}^{T} : = a_{j i}$ satisfies (H): its $i$ -th row total is $\sum_{j} ∣ a_{j i} ∣$ , which converges because its partial sums $\sum_{j < Q} ∣ a_{j i} ∣$ are bounded by $L$ by step 1.1; and the partial sums $\sum_{i < P} \sum_{j} ∣ a_{j i} ∣$ are limits of the rectangle sums $\sum_{i < P} \sum_{j < Q} ∣ a_{j i} ∣$ , again bounded by $L$ by step 1.1, so the series of row totals converges.

step 1.1L1L2L10

3.1

For every $N$ , $∣ S - \sum_{n < N} a_{J (n)} ∣ \leq Λ - \sum_{n < N} ∣ a_{J (n)} ∣$ : for $M > N$ the triangle inequality gives $∣ \sum_{n < M} a_{J (n)} - \sum_{n < N} a_{J (n)} ∣ \leq \sum_{n < M} ∣ a_{J (n)} ∣ - \sum_{n < N} ∣ a_{J (n)} ∣ \leq Λ - \sum_{n < N} ∣ a_{J (n)} ∣$ , and letting $M$ grow, the limit preserves the two non-strict inequalities bounding the left side.

step 2.2L1L4L10

3.2

Let $a^{'}$ agree with $a$ on the rectangle ${(i, j) : i < P_{0}, j < Q_{0}}$ and vanish off it. By step 2.1, $D = \sum_{n < N} a_{J (n)}^{'}$ and $E = \sum_{n < N} ∣ a_{J (n)}^{'} ∣$ ; since $∣ a_{J (n)}^{'} ∣ \leq ∣ a_{J (n)} ∣$ termwise, monotonicity gives $E \leq \sum_{n < N} ∣ a_{J (n)} ∣ \leq Λ \leq L$ .

step 2.1step 2.2step 2.4L1L2

4.1

By step 3.1 and step 3.2, $∣ S - \sum_{n < N} a_{J (n)} ∣ \leq Λ - \sum_{n < N} ∣ a_{J (n)} ∣ \leq L - E < 2 ε$ .

step 3.1step 2.3step 3.2

4.2

Also $∣ \sum_{n < N} a_{J (n)} - D ∣ = ∣ \sum_{n < N} (a - a^{'})_{J (n)} ∣ \leq \sum_{n < N} ∣ (a - a^{'})_{J (n)} ∣ \leq \sum_{i < P} \sum_{j < Q} ∣ (a - a^{'})_{i j} ∣ = \sum_{i < P} \sum_{j < Q} ∣ a_{i j} ∣ - E \leq L - E < 2 ε$ , the middle inequality by step 1.3 and the following equality by splitting the iterated sum at $P_{0}$ and at $Q_{0}$ , the array $a - a^{'}$ agreeing with $a$ off the small rectangle and vanishing on it.

step 1.1step 2.1step 1.3step 2.3step 2.4step 3.2L1L4

4.3

For each $i < P_{0}$ , $∣ R_{i} - \sum_{j < Q_{0}} a_{i j} ∣ \leq A_{i} - \sum_{j < Q_{0}} ∣ a_{i j} ∣ < ε / P_{0}$ , by the argument of step 3.1 applied to the row $i$ ; summing over $i < P_{0}$ gives $∣ \sum_{i < P_{0}} R_{i} - D ∣ < ε$ .

step 3.1step 1.6L1L4

4.4

Writing $Σ R$ for the sum of $\sum_{i} R_{i}$ , the same argument applied to the series $\sum_{i} R_{i}$ and the comparison $∣ R_{i} ∣ \leq A_{i}$ gives $∣ Σ R - \sum_{i < P_{0}} R_{i} ∣ \leq \sum_{i = 0}^{\infty} ∣ R_{i} ∣ - \sum_{i < P_{0}} ∣ R_{i} ∣ \leq L - \sum_{i < P_{0}} A_{i} < ε$ .

step 3.1step 1.5step 1.6L1L2

5.1

Combining step 4.1, step 4.2, step 4.3 and step 4.4, $∣ Σ R - S ∣ < ε + ε + 2 ε + 2 ε = 6 ε$ . As $ε > 0$ was arbitrary and $∣ Σ R - S ∣ \geq 0$ , this forces $Σ R = S$ , which with step 1.5 is claim 2.

step 1.5step 4.1step 4.2step 4.3step 4.4L5

6.1

Applying claims 1 and 2 to $a^{T}$ and to the bijection $J^{T}$ obtained by exchanging the coordinates of $J$ gives claim 3, since $a_{J^{T} (n)}^{T} = a_{J (n)}$ for every $n$ , so the two linear series are the same series and have the same sum $S$ .

step 2.2step 5.1step 2.5L7∎

Remarks

What the finite bookkeeping of steps 1.2 to 1.5 does, and why it is proved. Three facts are needed and none of them is among the laws of Laws of finite sums and finite products, all of which compare sums term by term over the same index range: that a sum along an injective list picks up an isolated term exactly once; that an iterated sum over a rectangle equals the sum along any injective list containing that rectangle, of the array cut down to it; and that a sum of nonnegative terms along an injective list into a rectangle is at most the iterated sum over the rectangle. Each is proved by zeroing out one entry at a time, which keeps the argument inside those laws.
Where the hypothesis is used. Only through step 1.1, which bounds every rectangle by $L$ , and through step 1.6, which makes a single rectangle capture all but $2 ε$ of the total mass. Everything else is bookkeeping. This is why the hypothesis has to be an absolute one: for a signed array no rectangle captures the mass, and the two iterated sums can disagree.
The independence of the enumeration is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum and nothing more. Two bijections $N \to N \times N$ differ by a bijection of $N$ , and an absolutely convergent series is unconditionally convergent. So the "sum of the array" is a well-defined real number attached to the array itself, and the theorem says the two iterated sums compute it.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors

Definition

Let $(a_{k})$ be a sequence of reals. Its partial products are

$Π_{n} : = \prod_{k < n} a_{k} (n \in N),$

the finite products of Finite sums and finite products, by recursion, so that $Π_{0} = 1$ , the empty product, and $Π_{n + 1} = Π_{n} a_{n}$ . For $N \in N$ the $N$ -th tail products are $T_{n}^{(N)} : = \prod_{j < n} a_{N + j}$ , again a sequence in $n$ .

Convergence. The infinite product $\prod a_{k}$ converges when there exists $N \in N$ such that

$a_{k} \neq 0$ for every $k \geq N$ , and
the sequence $(T_{n}^{(N)})_{n}$ of $N$ -th tail products converges (Limits and Cauchy sequences of reals) to a limit $ℓ \neq 0$ .

Its value is then

$\prod_{k = 0}^{\infty} a_{k} : = (\prod_{k < N} a_{k}) \cdot ℓ .$

If no such $N$ exists, the product diverges.

The value does not depend on $N$ , and that is a proof obligation, discharged here. First, if $N$ is such an index then so is every $N^{'} \geq N$ : condition 1 is inherited, and splitting the finite product (Laws of finite sums and finite products) gives, for $n \geq N^{'} - N$ ,

$T_{n}^{(N)} = (\prod_{k = N}^{N^{'} - 1} a_{k}) T_{n - (N^{'} - N)}^{(N^{'})},$

where the bracketed factor is a product of finitely many nonzero reals and so is itself nonzero. Hence $(T_{m}^{(N^{'})})_{m}$ converges, to $ℓ^{'} = ℓ / \prod_{k = N}^{N^{'} - 1} a_{k}$ by the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients), and $ℓ^{'} \neq 0$ because $ℓ \neq 0$ . Second, the two candidate values agree:

$(\prod_{k < N^{'}} a_{k}) ℓ^{'} = (\prod_{k < N} a_{k}) (\prod_{k = N}^{N^{'} - 1} a_{k}) \frac{ℓ}{\prod_{k = N}^{N^{'} - 1} a_{k}} = (\prod_{k < N} a_{k}) ℓ,$

again by splitting. Finally, any two admissible indices $N_{1}, N_{2}$ are both at most $\max {N_{1}, N_{2}}$ , which is therefore admissible and gives the same value as each. Since a convergent sequence has exactly one limit (A sequence has at most one limit), the displayed value is a single well-determined real number.

Why a zero limit is excluded. The definition demands $ℓ \neq 0$ , not merely that the tail products converge. Both parts of the definition are doing work, and against different naive alternatives. Against the naive " $Π_{n}$ converges", with no tail clause at all: every sequence with a single zero factor has all its partial products equal to $0$ from that index on, hence convergent to $0$ , so "the product converges" would say nothing whatever about the factors — which is what condition 1, the restriction to a tail of nonzero factors, repairs. Against the naive "some tail of the partial products converges", which keeps condition 1 and drops only $ℓ \neq 0$ , condition 1 no longer helps, and a product like $\prod_{j \geq 0} (1 - 1 / (j + 2))$ , all of whose factors are nonzero, has partial products $1 / (n + 1)$ tending to $0$ ; calling that convergent would make the value $0$ without any factor being $0$ , and would destroy the analogy with series in which a convergent product may be divided by. That product is worked out on the companion examples page.

Remarks

The value is $0$ exactly when some factor is $0$ . With $N$ as in the definition, the value is $(\prod_{k < N} a_{k}) ℓ$ with $ℓ \neq 0$ , and a finite product vanishes exactly when one of its factors does — a field has no zero divisors (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ), so an induction on the recursion $\prod_{k < n + 1} a_{k} = (\prod_{k < n} a_{k}) a_{n}$ of Finite sums and finite products, by recursion gives both directions. So a convergent product with all factors nonzero has nonzero value, and this is the property that makes convergent products behave like nonzero numbers.
Finitely many factors may be $0$ , or negative, or anything at all. The definition looks only at a tail, exactly as Series, partial sums, convergence and the sum, divergence, and the tail series does for series through its tail clause; conditions 1 and 2 constrain no initial segment.
Notation. $\prod_{k \geq m} a_{k}$ denotes the product of the family from $m$ , that is the product of the sequence $j \mapsto a_{j + m}$ , by the same convention Series, partial sums, convergence and the sum, divergence, and the tail series uses for series; the two readings agree at $m = 0$ .
Nothing here presumes a logarithm. The classical criteria for infinite products are usually derived by taking logarithms; the logarithm is not available at this point in the reading order, and For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent is proved from elementary inequalities instead.

TheoremStatement: AI-adaptedProof: AI-adaptedverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent

Statement

Write $S_{n} : = \sum_{k < n} p_{k}$ and $Π_{n} : = \prod_{k < n} (1 + p_{k})$ , $Q_{n} : = \prod_{k < n} (1 - p_{k})$ (Finite sums and finite products, by recursion, Series, partial sums, convergence and the sum, divergence, and the tail series, Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Elementary inequalities. Let $p_{k} \geq 0$ for every $k$ . Then for every $n \in N$ : $1 + S_{n} \leq Π_{n}, and Π_{n} \leq \frac{1}{1 - S_{n}} whenever S_{n} < 1;$ and if in addition $p_{k} \leq 1$ for every $k$ , then $1 - S_{n} \leq Q_{n} and Q_{n} Π_{n} \leq 1, hence Q_{n} \leq \frac{1}{1 + S_{n}} .$
The nonnegative criterion. Let $p_{k} \geq 0$ for every $k$ . Then $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges.
The $(1 - p_{k})$ form. Let $0 \leq p_{k} < 1$ for every $k$ . Then $\prod (1 - p_{k})$ converges if and only if $\sum p_{k}$ converges; and if $\sum p_{k}$ diverges then $Q_{n} \to 0$ , so that no tail of the product has partial products with a nonzero limit.
Absolute convergence. Let $(p_{k})$ be an arbitrary sequence of reals with $\sum ∣ p_{k} ∣$ convergent. Then $\prod (1 + p_{k})$ converges.

No logarithm occurs anywhere. The exponential and the logarithm, through which these criteria are usually derived, are later in the reading order; every inequality above is an induction on finite products. The refinement that decides $\prod (1 + p_{k})$ for signed $p_{k}$ with $\sum p_{k}$ convergent, in terms of the convergence of $\sum p_{k}^{2}$ , does need the logarithm and is not stated here; see Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

Facts & Assumptions

Given: A sequence $(p_{k})$ of reals, with $S_{n} = \sum_{k < n} p_{k}$ , $Π_{n} = \prod_{k < n} (1 + p_{k})$ and $Q_{n} = \prod_{k < n} (1 - p_{k})$ .

[L1]

Finite sums and products: $\sum_{k < 0} x_{k} = 0$ , $\prod_{k < 0} x_{k} = 1$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , $\prod_{k < n + 1} x_{k} = (\prod_{k < n} x_{k}) x_{n}$ , splitting at an intermediate index, and $\prod_{k < n} (x_{k} y_{k}) = (\prod_{k < n} x_{k}) (\prod_{k < n} y_{k})$ ; a finite product of nonnegative factors is nonnegative and of positive factors is positive (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

The principle of induction on $N$ (The principle of mathematical induction).

[L3]

For a series of nonnegative terms: convergence is equivalent to the range of the partial sums being bounded above, the sum is then the supremum and every partial sum is at most the sum, and if the range is unbounded the partial sums diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Divergence to $+ \infty$ and to $- \infty$ ).

[L4]

A series converges if and only if some tail series converges, and then the sum equals the initial partial sum plus the tail sum (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L5]

A nondecreasing sequence bounded above converges, and a nonincreasing sequence bounded below converges; a monotone sequence converges if and only if it is bounded (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, A monotone sequence converges if and only if it is bounded, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L6]

Order and inverses: $0 < a < b$ implies $0 < 1 / b < 1 / a$ , and $a > 0$ implies $1 / a > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[L7]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , $- ∣ x ∣ \leq x \leq ∣ x ∣$ , and $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ (Basic properties of the absolute value).

[L8]

[L9]

The squeeze theorem (The squeeze theorem).

[L10]

Every Cauchy sequence of reals converges (The reals are complete, Limits and Cauchy sequences of reals).

[L11]

Convergence of an infinite product: some tail has nonvanishing factors and partial products with a nonzero limit (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Proof

technique · direct

1.1

Assume $p_{k} \geq 0$ for every $k$ . An induction gives $1 + S_{n} \leq Π_{n}$ : at $n = 0$ both sides are $1$ ; and if $1 + S_{n} \leq Π_{n}$ then, since $1 + S_{n} \geq 1 > 0$ and $1 + p_{n} \geq 1 > 0$ , $Π_{n + 1} = Π_{n} (1 + p_{n}) \geq (1 + S_{n}) (1 + p_{n}) = 1 + S_{n} + p_{n} + S_{n} p_{n} \geq 1 + S_{n + 1}$ .

givenL1L2

1.2

Assume further $p_{k} \leq 1$ for every $k$ . An induction gives $1 - S_{n} \leq Q_{n}$ : at $n = 0$ both sides are $1$ ; and if $1 - S_{n} \leq Q_{n}$ then, since $1 - p_{n} \geq 0$ , $Q_{n + 1} = Q_{n} (1 - p_{n}) \geq (1 - S_{n}) (1 - p_{n}) = 1 - S_{n} - p_{n} + S_{n} p_{n} \geq 1 - S_{n + 1}$ .

givenL1L2

1.3

Assume $p_{k} \geq 0$ and $\sum p_{k}$ convergent, with sum $L$ . By [L3] and [L4] the tail sums $L - S_{N}$ tend to $0$ , so fix $N$ with $\sum_{k \geq N} p_{k} < 1 / 2$ .

givenL3L4choose

1.4

Three inductions on finite products, valid for arbitrary reals $x_{k}, y_{k}, z_{k}$ : first, $∣ \prod_{k < n} x_{k} ∣ = \prod_{k < n} ∣ x_{k} ∣$ , from $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ 1 ∣ = 1$ ; second, if $0 \leq x_{k} \leq y_{k}$ for all $k < n$ then $\prod_{k < n} x_{k} \leq \prod_{k < n} y_{k}$ , since the products are nonnegative and $\prod_{k < n + 1} x_{k} = (\prod_{k < n} x_{k}) x_{n} \leq (\prod_{k < n} y_{k}) x_{n} \leq (\prod_{k < n} y_{k}) y_{n}$ ; third, $∣ \prod_{k < n} (1 + z_{k}) - 1 ∣ \leq \prod_{k < n} (1 + ∣ z_{k} ∣) - 1$ , since at $n = 0$ both sides are $0$ and $∣ \prod_{k < n + 1} (1 + z_{k}) - 1 ∣ = ∣ (\prod_{k < n} (1 + z_{k}) - 1) (1 + z_{n}) + z_{n} ∣ \leq (\prod_{k < n} (1 + ∣ z_{k} ∣) - 1) (1 + ∣ z_{n} ∣) + ∣ z_{n} ∣ = \prod_{k < n + 1} (1 + ∣ z_{k} ∣) - 1$ .

L1L2L7

1.5

Assume $\sum ∣ p_{k} ∣$ converges, with sum $L$ , and fix $N$ with $τ_{N} : = \sum_{k \geq N} ∣ p_{k} ∣ < 1 / 2$ ; write $τ_{N + n} = \sum_{k \geq N + n} ∣ p_{k} ∣$ , so $τ_{N + n} \to 0$ and $\sum_{j < m} ∣ p_{N + j} ∣ \leq τ_{N}$ for every $m$ . For $k \geq N$ we get $∣ p_{k} ∣ \leq τ_{N} < 1 / 2$ , so $1 + p_{k} \geq 1 / 2 > 0$ and every factor from $N$ on is nonzero.

givenL3L4L7choose

2.1

An induction gives: for every $n$ with $S_{n} < 1$ , $Π_{n} (1 - S_{n}) \leq 1$ . At $n = 0$ this reads $1 \cdot 1 \leq 1$ . Suppose it holds at $n$ and $S_{n + 1} < 1$ ; then $S_{n} \leq S_{n + 1} < 1$ , so $Π_{n} \leq 1 / (1 - S_{n})$ by [L6], and $Π_{n + 1} = Π_{n} (1 + p_{n}) \leq (1 + p_{n}) / (1 - S_{n})$ . Multiplying out, $(1 + p_{n}) (1 - S_{n} - p_{n}) = 1 - S_{n} - p_{n} S_{n} - p_{n}^{2} \leq 1 - S_{n}$ , and dividing by the positive $(1 - S_{n}) (1 - S_{n + 1})$ turns this into $(1 + p_{n}) / (1 - S_{n}) \leq 1 / (1 - S_{n + 1})$ .

givenstep 1.1L1L2L6

2.2

Under the same assumption, $Q_{n} Π_{n} = \prod_{k < n} (1 - p_{k}) (1 + p_{k}) = \prod_{k < n} (1 - p_{k}^{2}) \leq 1$ , the last step by the induction: the empty product is $1$ , and multiplying a value in $[0, 1]$ by a factor $1 - p_{n}^{2} \in [0, 1]$ again gives a value in $[0, 1]$ . Since $Π_{n} \geq 1 + S_{n} \geq 1 > 0$ , dividing gives $Q_{n} \leq 1 / Π_{n} \leq 1 / (1 + S_{n})$ . This completes claim 1.

step 1.1step 1.2L1L2L6

2.3

Assume $0 \leq p_{k} < 1$ and $\sum p_{k}$ convergent. Fix $N$ with $\sum_{k \geq N} p_{k} < 1 / 2$ as in step 1.3. By step 1.2 applied to the shifted sequence, $U_{n} : = \prod_{j < n} (1 - p_{N + j}) \geq 1 - \sum_{j < n} p_{N + j} \geq 1 / 2$ for every $n$ ; and $(U_{n})$ is nonincreasing, each factor lying in $(0, 1]$ . So $(U_{n})$ converges to a limit $\geq 1 / 2 > 0$ , and every factor $1 - p_{k}$ is positive, hence nonzero; $\prod (1 - p_{k})$ converges.

step 1.2step 1.3L1L5L8L11

3.1

For the shifted sequence $j \mapsto p_{N + j}$ , whose partial sums are at most $1 / 2 < 1$ , step 2.1 gives $T_{n} : = \prod_{j < n} (1 + p_{N + j}) \leq 1 / (1 - 1 / 2) = 2$ for every $n$ , and step 1.1 gives $T_{n} \geq 1$ . The sequence $(T_{n})$ is nondecreasing, each factor being at least $1$ , so it converges to a limit $ℓ$ with $1 \leq ℓ \leq 2$ ; in particular $ℓ \neq 0$ , and every factor $1 + p_{k}$ is at least $1$ , hence nonzero. So $\prod (1 + p_{k})$ converges.

step 1.1step 2.1step 1.3L1L5L8L11

3.2

Assume instead $0 \leq p_{k} < 1$ and $\sum p_{k}$ divergent. Then $S_{n} \to + \infty$ by [L3], so given a real $ε > 0$ there is $K$ with $S_{n} > 1 / ε$ for $n \geq K$ , whence $0 < 1 / (1 + S_{n}) < ε$ ; thus $1 / (1 + S_{n}) \to 0$ . By step 2.2, $0 \leq Q_{n} \leq 1 / (1 + S_{n})$ , so $Q_{n} \to 0$ by the squeeze.

step 2.2L3L6L9

3.3

Put $T_{n} : = \prod_{j < n} (1 + p_{N + j})$ . By step 1.4 and step 2.1 applied to the nonnegative sequence $j \mapsto ∣ p_{N + j} ∣$ , $∣ T_{n} ∣ \leq \prod_{j < n} (1 + ∣ p_{N + j} ∣) \leq 1 / (1 - τ_{N}) \leq 2$ ; and by step 1.4 and step 1.2, $T_{n} \geq \prod_{j < n} (1 - ∣ p_{N + j} ∣) \geq 1 - τ_{N} \geq 1 / 2$ , each factor $1 + p_{N + j} \geq 1 - ∣ p_{N + j} ∣ \geq 0$ .

step 2.1step 1.2step 1.4step 1.5L1L7

4.1

Conversely assume $p_{k} \geq 0$ and $\prod (1 + p_{k})$ convergent, with $N$ as in [L11]. Since $Π_{n} = (\prod_{k < N} (1 + p_{k})) T_{n - N}$ for $n \geq N$ and $(T_{m})$ converges, the sequence $(Π_{n})$ converges, hence is bounded, say $Π_{n} \leq M$ for all $n$ . By step 1.1, $1 + S_{n} \leq M$ for every $n$ , so the partial sums of the nonnegative series $\sum p_{k}$ are bounded above and $\sum p_{k}$ converges. Claim 2 is step 3.1 together with this.

step 1.1step 3.1L1L3L5L11

4.2

In that situation the product diverges: for any $N$ , $Q_{N + n} = (\prod_{k < N} (1 - p_{k})) U_{n}$ with $\prod_{k < N} (1 - p_{k}) > 0$ , so $U_{n} = Q_{N + n} / \prod_{k < N} (1 - p_{k}) \to 0$ and no tail has partial products with a nonzero limit. With step 2.3 this proves claim 3.

step 2.3step 3.2L1L8L11

4.3

For $m > n$ , splitting the product gives $T_{m} = T_{n} \prod_{j = n}^{m - 1} (1 + p_{N + j})$ , so $∣ T_{m} - T_{n} ∣ = ∣ T_{n} ∣ ∣ \prod_{j = n}^{m - 1} (1 + p_{N + j}) - 1 ∣ \leq 2 (\prod_{j = n}^{m - 1} (1 + ∣ p_{N + j} ∣) - 1) \leq 2 (\frac{1}{1 - τ_{N + n}} - 1) = \frac{2 τ_{N + n}}{1 - τ_{N + n}} \leq 4 τ_{N + n}$ , using step 2.1 for the shifted sequence from $N + n$ , whose partial sums are at most $τ_{N + n} \leq τ_{N} < 1 / 2$ .

step 2.1step 1.4step 1.5step 3.3L1L6L7

5.1

Since $τ_{N + n} \to 0$ , step 4.3 makes $(T_{n})$ a Cauchy sequence, so it converges, to a limit $ℓ$ ; and $ℓ \geq 1 / 2 > 0$ by step 3.3 and [L8]. Hence $\prod (1 + p_{k})$ converges, which is claim 4.

step 1.5step 3.3step 4.3L8L10L11∎

Remarks

Why the two bounds of claim 1 are the right pair. The lower bound $1 + S_{n} \leq Π_{n}$ is the Weierstrass product inequality and forces divergence of the product when $\sum p_{k}$ diverges; the upper bound $Π_{n} \leq 1 / (1 - S_{n})$ , available once the partial sums are below $1$ , forces convergence when $\sum p_{k}$ converges. Between them they prove claim 2 with no further input, and they are exactly what a logarithm would otherwise supply.
The strict inequality $p_{k} < 1$ keeps this proof uniform, but the tail-based definition allows a slightly stronger statement. Claim 3 remains true for $0 \leq p_{k} \leq 1$ . If only finitely many $p_{k}$ equal $1$ , start the product after the last zero factor; if infinitely many do, then $\sum p_{k}$ diverges and no tail has all factors nonzero. The stated strict form avoids this finite/infinite split.
Claim 4 does not identify the value, and the converse fails. Absolute convergence of $\sum p_{k}$ gives convergence of $\prod (1 + p_{k})$ , but convergence of $\sum p_{k}$ alone does not: the companion examples page exhibits $\sum_{j \geq 0} (- 1)^{j} / \sqrt{j + 2}$ convergent while the corresponding partial products tend to $0$ . What separates the two cases is the convergence of $\sum p_{k}^{2}$ , a criterion that needs the logarithm and is deferred.
Where the Cauchy criterion enters and why nothing cheaper would do. In claim 4 the factors have no sign, so the partial products are not monotone and A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum is unavailable; the estimate of step 4.3 is a Cauchy estimate and is closed by completeness of $R$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Base- $b$ expansions: for an integer $b \geq 2$ every $x \in [0, 1)$ is the sum of $\sum_{j \geq 0} d_{j} / b^{j + 1}$ for digits $d_{j} < b$ , and the digit sequence is unique among those that are not eventually constantly $b - 1$

Statement

Let $b \in N$ with $b \geq 2$ and write $β : = ι (b)$ for the canonical natural of $b$ in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field), so that $β > 1$ (Canonical naturals are positive and strictly increasing); powers $β^{n}$ are integer powers (Integer powers $a^{m}$ ). Call a sequence $(d_{j})$ of natural numbers a digit sequence in base $b$ when $d_{j} < b$ for every $j$ , and say it is terminal when it is eventually constantly $b - 1$ , that is when there is $J$ with $d_{j} = b - 1$ for every $j \geq J$ . Then:

Existence. For every $x \in [0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) there is a non-terminal digit sequence $(d_{j})$ in base $b$ with $\sum_{j \geq 0} \frac{ι (d_{j})}{β^{j + 1}} convergent, of sum x .$
Uniqueness. If $(c_{j})$ and $(c_{j}^{'})$ are non-terminal digit sequences in base $b$ whose series have the same sum, then $c_{j} = c_{j}^{'}$ for every $j$ .

So every real in $[0, 1)$ has exactly one base- $b$ expansion once the terminal sequences are excluded. The assignment $x \mapsto (d_{j})$ is moreover a bijection onto the non-terminal digit sequences: claim 2 makes it injective, and it is onto because a non-terminal $(c_{j})$ has $ι (c_{j}) \leq β - 1$ for every $j$ and $ι (c_{j}) < β - 1$ for infinitely many $j$ , so its sum is strictly below the sum $1$ of the all- $(b - 1)$ series computed in step 8.1, hence lies in $[0, 1)$ and has $(c_{j})$ as its expansion by claim 2. Excluding them is unavoidable: the terminal sequences are exactly the ones producing a second expansion of a number that already has one, as the companion examples page exhibits with $0.999 \dots = 1$ and $0.4999 \dots = 0.5$ .

The construction uses no floor function. The integer part of a real is not available at this point in the reading order, so the digit at each stage is produced by the finite case distinction "in which of the $b$ intervals $[d / β, (d + 1) / β)$ does the current residue lie", closed by the well-ordering principle (The well-ordering principle), and the digits are assembled by the recursion theorem (The recursion theorem).

Indices run from $0$ . The digit $d_{j}$ carries the weight $β^{- (j + 1)}$ , so that the first digit has weight $1 / β$ and no denominator $β^{0} = 1$ ever occurs.

Facts & Assumptions

Given: A natural number $b \geq 2$ , $β = ι (b)$ , and a real $x \in [0, 1)$ .

[L1]

The canonical natural: $ι (0) = 0$ , $ι (n + 1) = ι (n) + 1$ , $ι$ is strictly increasing on $N$ , and $ι (m + n) = ι (m) + ι (n)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L2]

Integer powers: $β^{0} = 1$ , $β^{n + 1} = β^{n} β$ , $(u v)^{n} = u^{n} v^{n}$ , and $β^{n} > 0$ for $β > 0$ (Integer powers $a^{m}$ , Laws of integer exponents).

[L3]

$[0, 1) = {y \in R : 0 \leq y < 1}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

The recursion theorem (The recursion theorem) and the principle of induction (The principle of mathematical induction).

[L6]

Geometric series: for $∣ r ∣ < 1$ , $\sum r^{k}$ converges with sum $1 / (1 - r)$ ; and the terms of a convergent series tend to $0$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, If a series converges then its terms tend to $0$ ).

[L7]

Order and inverses: $0 < u < v$ implies $0 < 1 / v < 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[L8]

Finite sums: recursion, splitting, additivity, scaling and monotonicity; in particular every single term of a finite sum of nonnegative reals is at most that sum (Laws of finite sums and finite products).

[L9]

Partial sums and sums of series; linearity of convergent series; and a series converges if and only if some tail series converges, the sum being the initial partial sum plus the tail sum (Series, partial sums, convergence and the sum, divergence, and the tail series, Convergent series add and scale termwise, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Limits and Cauchy sequences of reals).

[L10]

For a series of nonnegative terms, every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L11]

The squeeze theorem, the algebra of limits, and that limits preserve non-strict inequalities (The squeeze theorem, Algebra of limits: sums, scalar multiples, products and quotients, Limits preserve non-strict inequalities).

Proof

technique · constructive

1.1

$β = ι (b) \geq ι (2) = 1 + 1 > 1 > 0$ , and an induction gives $β^{n} > 0$ for every $n$ ; also $0 < 1 / β < 1$ .

givenL1L2L5L7

1.2

For uniqueness, let $(c_{j})$ and $(c_{j}^{'})$ be non-terminal digit sequences whose series have the same sum, and suppose they are not equal. By [L4] let $n$ be least with $c_{n} \neq c_{n}^{'}$ ; interchanging the two sequences if necessary, assume $c_{n} < c_{n}^{'}$ , so $ι (c_{n}^{'}) - ι (c_{n}) \geq 1$ .

givenL1L4choose

2.1

The digit of a residue. For every $r \in [0, 1)$ there is exactly one natural $d < b$ with $ι (d) / β \leq r < ι (d + 1) / β$ . For existence, the set $D : = {d \in N : d \leq b and r < ι (d) / β}$ contains $b$ , since $ι (b) / β = 1 > r$ ; let $m : = \min D$ , which is not $0$ because $ι (0) / β = 0 \leq r$ ; put $d : = m - 1$ , so $d < b$ , and minimality of $m$ says $d \notin D$ , that is $ι (d) / β \leq r$ , while $m \in D$ says $r < ι (m) / β = ι (d + 1) / β$ . For uniqueness, if $d < d^{'}$ both worked then $r < ι (d + 1) / β \leq ι (d^{'}) / β \leq r$ , which the order forbids. Write $d (r)$ for this digit.

step 1.1L1L3L4construct

2.2

Both series converge, by hypothesis, so by linearity $\sum_{j} (ι (c_{j}^{'}) - ι (c_{j})) / β^{j + 1}$ converges with sum $0$ ; its first $n$ terms vanish, so by [L9] the tail from $n$ also has sum $0$ , that is $0 = (ι (c_{n}^{'}) - ι (c_{n})) / β^{n + 1} + \sum_{j > n} (ι (c_{j}^{'}) - ι (c_{j})) / β^{j + 1}$ .

givenstep 1.2L9

3.1

The residue map. For $r \in [0, 1)$ put $f (r) : = β r - ι (d (r))$ . Multiplying $ι (d (r)) / β \leq r < ι (d (r) + 1) / β = (ι (d (r)) + 1) / β$ by $β > 0$ gives $ι (d (r)) \leq β r < ι (d (r)) + 1$ , so $0 \leq f (r) < 1$ ; thus $f$ is a function from $[0, 1)$ to $[0, 1)$ .

step 1.1step 2.1L1L3construct

3.2

Every difference satisfies $ι (c_{j}^{'}) - ι (c_{j}) \geq - (β - 1)$ , the digits lying in ${0, \dots, b - 1}$ , and $\sum_{j > n} (β - 1) / β^{j + 1} = \frac{β - 1}{β^{n + 2}} \cdot \frac{1}{1 - 1 / β} = \frac{1}{β^{n + 1}}$ by the geometric series; hence the tail in step 2.2 is at least $- 1 / β^{n + 1}$ .

step 1.2L1L2L6L11

4.1

By the recursion theorem applied to the set $[0, 1)$ , the element $x$ and the function $f$ , there is a unique sequence $(r_{n})$ in $[0, 1)$ with $r_{0} = x$ and $r_{n + 1} = f (r_{n})$ ; put $d_{j} : = d (r_{j})$ , a natural number $< b$ , so $(d_{j})$ is a digit sequence in base $b$ .

step 3.1L5construct

4.2

Write $A$ for the first summand and $B$ for the tail in step 2.2, so $A + B = 0$ while $A \geq 1 / β^{n + 1}$ by step 1.2 and $B \geq - 1 / β^{n + 1}$ by step 3.2; since the two lower bounds sum to $0$ , both must be attained, that is $A = 1 / β^{n + 1}$ and $B = - 1 / β^{n + 1}$ ; in particular $\sum_{j > n} (ι (c_{j}^{'}) - ι (c_{j}) + (β - 1)) / β^{j + 1} = 0$ , a convergent series of nonnegative terms with sum $0$ , so every term is $0$ and $ι (c_{j}^{'}) - ι (c_{j}) = - (β - 1)$ for every $j > n$ .

step 1.2step 2.2step 3.2L8L9L10

5.1

An induction gives $x = \sum_{j < n} ι (d_{j}) / β^{j + 1} + r_{n} / β^{n}$ for every $n$ : at $n = 0$ the sum is empty and $r_{0} / β^{0} = x$ ; and from $r_{n} = (ι (d_{n}) + r_{n + 1}) / β$ , which is step 3.1 rearranged, one gets $r_{n} / β^{n} = ι (d_{n}) / β^{n + 1} + r_{n + 1} / β^{n + 1}$ .

step 3.1step 4.1L2L5L8

5.2

Since $0 \leq r_{n} < 1$ and $β^{n} > 0$ , we have $0 \leq r_{n} / β^{n} \leq 1 / β^{n} = (1 / β)^{n}$ ; as $0 < 1 / β < 1$ the series $\sum (1 / β)^{k}$ converges, so $(1 / β)^{n} \to 0$ , and the squeeze gives $r_{n} / β^{n} \to 0$ .

step 1.1step 4.1L2L6L7L11

5.3

That forces $c_{j} = b - 1$ and $c_{j}^{'} = 0$ for every $j > n$ , since the difference of two digits attains $- (β - 1)$ only at those values; so $(c_{j})$ is terminal, contrary to hypothesis. Hence the two digit sequences agree, which is claim 2.

step 1.2step 4.2L1

6.1

By step 5.1 the partial sums of $\sum_{j} ι (d_{j}) / β^{j + 1}$ equal $x - r_{n} / β^{n}$ , which converges to $x$ ; so the series converges with sum $x$ .

step 5.1step 5.2L9L11

7.1

Applying step 5.1 and step 6.1 to the residue $r_{J}$ in place of $x$ , whose recursion produces the digits $d_{J + i}$ , gives $r_{J} = \sum_{i \geq 0} ι (d_{J + i}) / β^{i + 1}$ for every $J$ .

step 4.1step 6.1L5

8.1

The constructed sequence is not terminal: if $d_{j} = b - 1$ for every $j \geq J$ , then by step 7.1 and the geometric series, $r_{J} = \sum_{i \geq 0} (β - 1) / β^{i + 1} = \frac{β - 1}{β} \cdot \frac{1}{1 - 1 / β} = 1$ , contradicting $r_{J} < 1$ ; here $ι (b - 1) = β - 1$ by [L1]. With step 6.1 this proves claim 1.

step 6.1step 7.1L1L2L6L9

9.1

Claim 1 is step 6.1 with step 8.1 and claim 2 is step 5.3, so every $x \in [0, 1)$ has exactly one non-terminal base- $b$ expansion.

step 6.1step 8.1step 5.3discharge-construct∎

Remarks

Where each tool is used, and the floor function is not among them. The well-ordering principle appears once, in step 2.1, to pick out the digit from the finitely many candidates $0, \dots, b$ ; the recursion theorem appears once, in step 4.1, to turn the one-step residue map into a sequence. Everything else is the geometric series and the ordering of $R$ . The usual formula $d_{n} = ⌊ β r_{n} ⌋$ would need the integer part of a real, which is developed later in the reading order.
The exclusion of terminal sequences is exactly one equivalence class. Step 4.2 shows that two distinct expansions of the same number must differ by one at the first place where they differ and then be all $b - 1$ against all $0$ . So each real in $(0, 1)$ whose expansion terminates in zeros has exactly two expansions and every other real exactly one; forbidding the all- $(b - 1)$ tails picks one from each pair.
The hypothesis $x < 1$ is not a restriction on the theorem so much as on the notation. The all- $(b - 1)$ sequence sums to $1$ , as step 8.1 computes, and $1$ is not in $[0, 1)$ ; a base- $b$ expansion of a general nonnegative real is an integer part together with an expansion of the fractional part, and the integer part is not available here.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-29Open item page →

The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order

Remark

Everything on this page is about series of real numbers, and the answer it reaches is complete for that case. Write

$S (a) : = {s \in R : some rearrangement of \sum a_{k} converges to s}$

for the set of rearrangement sums of a convergent series (Rearrangement of a series along a bijection of $N$ , and unconditional convergence). Then this page determines $S (a)$ exactly, in two cases and no others.

If $\sum a_{k}$ converges absolutely, $S (a)$ is the single point ${\sum_{k = 0}^{\infty} a_{k}}$ : that is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum.
If $\sum a_{k}$ converges conditionally, $S (a)$ is the whole of $R$ : that is The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ , and moreover rearrangements exist whose partial sums diverge to $+ \infty$ or to $- \infty$ or oscillate between any prescribed pair of extended reals.

For a series of real numbers, unconditional convergence and absolute convergence are the same property is the statement that these two cases are distinguished by absolute convergence and by nothing else.

The same question can be asked of a series of vectors, once one has a space in which a series of vectors has a sum: given a convergent series in $R^{d}$ , what does its set of rearrangement sums look like? That question was raised by Paul Lévy in 1905 and taken up by Ernst Steinitz in 1913, and later by Wacław Sierpiński; the references below are to those papers, and they are given as the origin of the question. What the literature answers is not stated here in any form, and nothing on this page or anywhere else in this library depends on it. Part of it is now proved, later in the reading order and marked as forward material: Steinitz's polygonal confinement theorem: finitely many vectors of norm at most $1$ summing to $0$ can be ordered so that every partial sum has norm at most $n$ ↗ and The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$ ↗ establish that the set of rearrangement sums is nonempty and lies inside an affine subspace. The reverse inclusion, which is what would turn that containment into the classical answer, is still proved nowhere here.

The reason is a matter of reading order, not of difficulty or of interest. Stating the theorem requires $R^{d}$ as a normed space (a norm, convergence of vector sequences, and a notion of a convergent series of vectors), and that vocabulary is introduced later in the reading order than this page. Rather than borrow it, or state a theorem whose terms are not yet defined, the obligation is recorded where it can be discharged: on the page that builds $R^{d}$ as a normed space and afterwards. When that page is reached, the question raised here is the one it will answer.

What is safe to say now, and is worth saying. The one-dimensional dichotomy above is stark: a single point, or everything. Nothing in the proof of The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ survives verbatim in higher dimensions, because it is built on the order of $R$ : the greedy rule "add positive terms until the running sum exceeds the target, then negative ones until it falls below" presupposes that the terms are signed and that the target can be approached from two sides. In $R^{d}$ with $d \geq 2$ there is no such order, the terms point in many directions, and the argument has no analogue. A reader who expects the one-dimensional answer to generalise unchanged should treat that expectation as unsupported until the later page settles it.

No claim of this library is made about $R^{d}$ above. The two Lévy and Steinitz papers are cited as the historical source of the question, not as authority for a result used anywhere here; no item on this page or elsewhere in the library rests on them.

RemarkRemark: AI-adaptedProof: Not applicableverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for

Remark

A convergence test proves that a limit exists; it does not produce the limit. On this page that gap is systematic, and this remark records the principal places where a familiar value or formula is deferred and what would close it. Every scope statement below is relative to the reading order: the material named is developed elsewhere in this library, later than this page, and nothing here says it is absent from the library.

The alternating harmonic series. The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ proves that $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges, and its error bound pins the sum between consecutive partial sums; the companion examples page uses that to prove the sum lies strictly between $1 / 2$ and $1$ . No closed expression for the sum is given, and none can be given here: the classical value is a logarithm, and the logarithm is introduced later in the reading order. So the sum is named, bracketed, and left unevaluated.

The two-positive-one-negative rearrangement. The same is true one level up. The companion examples page proves that taking two positive terms for each negative one produces a convergent rearrangement whose sum is $3 / 2$ times the sum of the original series. That statement is exact and complete as it stands, and it is deliberately relative: it compares two sums rather than evaluating either. The familiar form of the same fact multiplies a logarithm by $3 / 2$ , and it becomes available at the same later point.

The refined criterion for infinite products. For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent settles $\prod (1 + p_{k})$ completely for $p_{k} \geq 0$ , settles $\prod (1 - p_{k})$ for $0 \leq p_{k} < 1$ , and proves that $\sum ∣ p_{k} ∣$ convergent forces $\prod (1 + p_{k})$ convergent. It does not settle the remaining case: a signed sequence $(p_{k})$ with $\sum p_{k}$ convergent but $\sum ∣ p_{k} ∣$ divergent. The classical criterion there is that $\prod (1 + p_{k})$ converges exactly when $\sum p_{k}^{2}$ converges. A standard proof expands $\log (1 + x)$ ; that route belongs with the logarithm, later in the reading order. The gap is not hypothetical: the companion examples page exhibits a signed sequence with $\sum p_{k}$ convergent whose partial products tend to $0$ .

Rearrangement beyond $R$ . The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ and For a series of real numbers, unconditional convergence and absolute convergence are the same property together answer the rearrangement question for real series completely. The corresponding question for series of vectors is raised, and left open at this point in the reading order, in The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order, which states no theorem about it.

Two places where existence is constructive but no formula is claimed. Base- $b$ expansions: for an integer $b \geq 2$ every $x \in [0, 1)$ is the sum of $\sum_{j \geq 0} d_{j} / b^{j + 1}$ for digits $d_{j} < b$ , and the digit sequence is unique among those that are not eventually constantly $b - 1$ produces, for every $x \in [0, 1)$ , its digit sequence in base $b$ , by a recursion that depends on $x$ ; it gives no closed expression for the digits of any particular real, and it claims none. Likewise The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ produces, for each prescribed target, a bijection of $N$ defined by a recursion over the terms of the series; no formula for that bijection is given, and the theorem asserts only that one exists. In both cases the construction is fully determined by the data, with no choice made anywhere, which is a stronger statement than mere existence and a weaker one than a formula.

What this list does not claim. It is not a census of every convergence result on the page. In particular, the Dirichlet, alternating-series, and Abel tests and their worked applications establish additional convergence without evaluating a numerical sum; their purpose here is to supply convergence criteria, not to flag a familiar value whose evaluation waits for a later object. Among the structural comparison theorems, Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ , Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum and Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value identify sums with one another and evaluate nothing, which is exactly what makes them usable wherever the sums themselves are unknown.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: every convergent series converges absolutely

Statement

False claim: for every sequence $(a_{k})$ of reals, if $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then $\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).

What is true is the converse, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges: absolute convergence implies convergence. The claim above reverses it, and the reversal fails at the standard witness, the alternating harmonic series.

$a_{j} : = \frac{ε_{j}}{ι (j + 1)} (j \in N),$

with $ι (j + 1)$ the canonical natural (Canonical naturals are positive and strictly increasing). Then $\sum a_{j}$ converges while $\sum ∣ a_{j} ∣$ is the harmonic series, which diverges. So the two notions really are different, and "conditionally convergent" is not an empty class.

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ , the sequence $b_{j} : = 1 / ι (j + 1)$ , and $a_{j} : = ε_{j} b_{j}$ .

[A1]

The refuted claim: every convergent series of reals converges absolutely.

[L1]

$∣ ε_{j} ∣ = 1$ for every $j$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

The canonical naturals $ι (n)$ are positive for $n \geq 1$ and strictly increasing in $n$ (Canonical naturals are positive and strictly increasing).

[L3]

If $0 < u < v$ then $0 < 1 / v < 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / ι (n) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L5]

The alternating series test: if $(b_{j})$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) with $b_{j} \to 0$ , then $\sum ε_{j} b_{j}$ converges (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , Limits and Cauchy sequences of reals).

[L6]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ , where $k^{p} = ι (k)^{p}$ ; at $p = 1$ the rational power is the element itself, $ι (k)^{1} = ι (k)$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ ).

[L7]

The series $\sum_{k \geq 1} x_{k}$ is by definition the series of the sequence $j \mapsto x_{j + 1}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L8]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ (Basic properties of the absolute value).

[L9]

Absolute convergence means convergence of $\sum ∣ a_{j} ∣$ ; conditional convergence means convergence of $\sum a_{j}$ without it (Absolutely convergent and conditionally convergent series, and the general starting index).

[L10]

Absolute convergence implies convergence (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Refutation

technique · direct

1.1

Each $b_{j} = 1 / ι (j + 1)$ is a positive real, $ι (j + 1)$ being a positive canonical natural.

givenL2

1.2

The sequence $(b_{j})$ is nonincreasing: $ι (j + 1) < ι (j + 2)$ , so $1 / ι (j + 2) < 1 / ι (j + 1)$ .

L2L3

2.1

The sequence $(b_{j})$ converges to $0$ : given a rational $ε > 0$ , fix a natural $n \geq 1$ with $1 / ι (n) < ε$ ; then for every $j \geq n$ one has $ι (j + 1) \geq ι (n) > 0$ , hence $∣ b_{j} ∣ = b_{j} \leq 1 / ι (n) < ε$ .

step 1.1L2L3L4

2.2

For every $j$ , $∣ a_{j} ∣ = ∣ ε_{j} ∣ ∣ b_{j} ∣ = b_{j} = 1 / ι (j + 1)$ .

step 1.1L1L8

3.1

By the alternating series test, $\sum a_{j} = \sum ε_{j} b_{j}$ converges.

step 1.2step 2.1L5

3.2

The series $\sum_{j} 1 / ι (j + 1)$ is, by the definition of a series from a general starting index, exactly the series $\sum_{k \geq 1} 1 / k$ , that is the $p$ -series at $p = 1$ .

step 2.2L6L7

4.1

The $p$ -series at $p = 1$ diverges, since $1 > 1$ is false; so $\sum ∣ a_{j} ∣$ diverges.

step 3.2L6

5.1

Thus $\sum a_{j}$ converges while $\sum ∣ a_{j} ∣$ does not, so $\sum a_{j}$ converges conditionally and not absolutely, and the claim [A1] fails for this series.

step 3.1step 4.1A1L9

6.1

The claim is therefore false. What survives of it is only the converse implication, that an absolutely convergent series converges.

step 5.1A1L10∎

Remarks

The witness is not exotic. It is the first series a reader meets whose convergence depends on cancellation, and the failure is as large as it can be: the series of absolute values does not merely converge to a different number, it diverges to $+ \infty$ .
Everything on this page turns on this example. Because the class of conditionally convergent series is nonempty, The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ has content, and For a series of real numbers, unconditional convergence and absolute convergence are the same property separates two genuinely different properties rather than restating one. The same series, with its rearrangements, is developed on the companion page.
The value of the sum is not asserted here. The series is proved convergent and nothing more; the classical evaluation needs the logarithm, which is later in the reading order (Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for).

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: every rearrangement of a convergent series converges, and to the same sum

Statement

False claim: for every sequence $(a_{k})$ of reals whose series converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and every bijection $σ : N \to N$ , the rearranged series $\sum a_{σ (k)}$ (Rearrangement of a series along a bijection of $N$ , and unconditional convergence) converges, with the same sum.

What is true is that hypothesis: the claim holds for absolutely convergent series, and that is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum. Dropping "absolutely" makes it false in both of its assertions at once, and the same witness refutes both.

Let $(ε_{j})$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ) and put $a_{j} : = ε_{j} / ι (j + 1)$ , the alternating harmonic series. It converges, by the alternating series test, and does not converge absolutely, its series of absolute values being the harmonic series (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). So it converges conditionally, and The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ applies to it.

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ , the sequence $b_{j} : = 1 / ι (j + 1)$ , and $a_{j} : = ε_{j} b_{j}$ , whose series is the alternating harmonic series.

[A1]

The refuted claim: for every convergent series of reals and every bijection of $N$ , the rearranged series converges with the same sum.

[L1]

[L2]

The canonical naturals $ι (n)$ are positive for $n \geq 1$ and strictly increasing; if $0 < u < v$ then $0 < 1 / v < 1 / u$ ; and for every real $ε > 0$ there is $n \geq 1$ with $1 / ι (n) < ε$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L3]

The alternating series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Limits and Cauchy sequences of reals).

[L4]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ , with $ι (k)^{1} = ι (k)$ ; and $\sum_{k \geq 1} x_{k}$ is the series of $j \mapsto x_{j + 1}$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ (Basic properties of the absolute value).

[L6]

Absolute and conditional convergence (Absolutely convergent and conditionally convergent series, and the general starting index).

[L7]

The Riemann series theorem: a conditionally convergent series has, for every real $c$ , a rearrangement converging to $c$ , and one whose partial sums diverge to $+ \infty$ (The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ , Divergence to $+ \infty$ and to $- \infty$ ).

[L8]

An absolutely convergent series converges unconditionally (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum).

Refutation

technique · direct

1.1

The sequence $(b_{j})$ is positive, nonincreasing and converges to $0$ : positivity and monotonicity from $0 < ι (j + 1) < ι (j + 2)$ , and convergence because, given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε$ satisfies $b_{j} \leq 1 / ι (n) < ε$ for every $j \geq n$ .

givenL2

2.1

By the alternating series test $\sum a_{j}$ converges; write $S$ for its sum.

step 1.1L3

2.2

For every $j$ , $∣ a_{j} ∣ = ∣ ε_{j} ∣ b_{j} = 1 / ι (j + 1)$ , and $\sum_{j} 1 / ι (j + 1)$ is the $p$ -series $\sum_{k \geq 1} 1 / k$ at $p = 1$ , which diverges.

step 1.1L1L4L5

3.1

So $\sum a_{j}$ converges conditionally.

step 2.1step 2.2L6

4.1

By the Riemann series theorem there is a bijection $σ$ of $N$ with $\sum a_{σ (k)}$ convergent of sum $S + 1$ , a number different from $S$ .

step 3.1L7

4.2

By the same theorem there is a bijection $τ$ of $N$ for which the partial sums of $\sum a_{τ (k)}$ diverge to $+ \infty$ , so that rearranged series does not converge at all.

step 3.1L7

5.1

The claim [A1] therefore fails twice over for the alternating harmonic series: once in its assertion that the sum is preserved, by step 4.1, and once in its assertion that the rearranged series converges, by step 4.2.

step 4.1step 4.2A1

6.1

The claim is false. What is true is the same statement with "converges" strengthened to "converges absolutely" in the hypothesis.

step 5.1A1L8∎

Remarks

Neither half of the claim survives. It is often stated as though the only risk were a change of value; step 4.2 shows the rearranged series may fail to converge, and The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ shows the partial sums may be made to oscillate between any two prescribed extended reals.
The hypothesis that repairs the claim is exactly the right one. By For a series of real numbers, unconditional convergence and absolute convergence are the same property, absolute convergence is not merely sufficient for the conclusion but necessary: a convergent series all of whose rearrangements converge is absolutely convergent. So there is no intermediate hypothesis to look for.
What is fixed and what is not. The terms of the series are fixed; only the order changes. That an infinite sum should depend on the order at all is the point of the example, and it is why Series, partial sums, convergence and the sum, divergence, and the tail series defines the sum as the limit of the partial sums of a sequence, not as a sum over a set of indices.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: the Cauchy product of two convergent series converges

Statement

False claim: if $\sum a_{k}$ and $\sum b_{k}$ both converge (Series, partial sums, convergence and the sum, divergence, and the tail series) then their Cauchy product $\sum c_{n}$ converges (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

What is true is Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , which requires one of the two factors to converge absolutely. Convergence of both is not enough, and the standard witness is a single series multiplied by itself.

Let $(ε_{k})$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ) and put

$a_{k} = b_{k} : = \frac{ε_{k}}{\sqrt{ι (k + 1)}} (k \in N),$

with $\sqrt{}$ the nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ) and $ι (k + 1)$ the canonical natural, positive for every $k$ (Canonical naturals are positive and strictly increasing). Then $\sum a_{k}$ converges, by the alternating series test, while the Cauchy product satisfies

$∣ c_{n} ∣ \geq \frac{2 ι (n + 1)}{ι (n + 2)} \geq 1 for every n \in N,$

so $(c_{n})$ does not converge to $0$ and $\sum c_{n}$ diverges (If a series converges then its terms tend to $0$ ).

Facts & Assumptions

Given: The alternating sequence $(ε_{k})$ , the sequence $β_{k} : = 1 / \sqrt{ι (k + 1)}$ , the sequence $a_{k} = b_{k} = ε_{k} β_{k}$ , and their Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[A1]

The refuted claim: the Cauchy product of two convergent series of reals converges.

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{k + 1} = - ε_{k}$ , and $∣ ε_{k} ∣ = 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Square roots: every $t \geq 0$ has a unique $\sqrt{t} \geq 0$ with $(\sqrt{t})^{2} = t$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L3]

The canonical naturals: $ι (n) > 0$ for $n \geq 1$ , $ι$ is strictly increasing, and $ι (m + n) = ι (m) + ι (n)$ (Canonical naturals are positive and strictly increasing).

[L4]

If $0 < u < v$ then $0 < 1 / v < 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / ι (n) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

[L7]

AM-GM for two nonnegative reals, in the product form: $u v \leq ((u + v) / 2)^{2}$ (The arithmetic mean, geometric mean inequality).

[L8]

Finite sums: the sum of a constant, monotonicity in the terms, and $\sum_{k = 0}^{n} x_{k} = \sum_{k < n + 1} x_{k}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L9]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[L10]

If $\sum x_{n}$ converges then $x_{n} \to 0$ (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L11]

The principle of induction on $N$ (The principle of mathematical induction).

[L12]

Mertens' theorem, which requires one factor to converge absolutely (Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ ).

Refutation

technique · direct

1.1

Square roots are strictly increasing on the nonnegative reals: if $0 \leq u < v$ and $\sqrt{u} \geq \sqrt{v}$ then $u = (\sqrt{u})^{2} \geq (\sqrt{v})^{2} = v$ , which is false; so $\sqrt{u} < \sqrt{v}$ . Also $\sqrt{u v} = \sqrt{u} \sqrt{v}$ for $u, v \geq 0$ , since $(\sqrt{u} \sqrt{v})^{2} = u v$ and $\sqrt{u} \sqrt{v} \geq 0$ , and $\sqrt{t^{2}} = t$ for $t \geq 0$ .

1.2

An induction on $j$ gives $ε_{m} ε_{j} = ε_{m + j}$ for all $m, j$ : at $j = 0$ this is $ε_{m} \cdot 1 = ε_{m}$ , and $ε_{m} ε_{j + 1} = ε_{m} (- ε_{j}) = - ε_{m + j} = ε_{m + j + 1}$ .

L1L11

2.1

Each $β_{k} = 1 / \sqrt{ι (k + 1)}$ is a positive real, and $(β_{k})$ is nonincreasing, since $0 < ι (k + 1) < ι (k + 2)$ gives $0 < \sqrt{ι (k + 1)} < \sqrt{ι (k + 2)}$ and inverting reverses the inequality.

step 1.1L3L4

2.2

$(β_{k})$ converges to $0$ : given a rational $ε > 0$ , fix a natural $n \geq 1$ with $1 / ι (n) < ε^{2}$ ; then for $k \geq n$ one has $ι (k + 1) \geq ι (n) > 1 / ε^{2} = (1 / ε)^{2}$ , so $\sqrt{ι (k + 1)} > 1 / ε$ and $β_{k} < ε$ .

step 1.1L3L4L5

2.3

For $k \leq n$ , [L7] applied to $u = ι (k + 1)$ and $v = ι (n - k + 1)$ , whose sum is $ι (n + 2)$ by [L3], gives $ι (k + 1) ι (n - k + 1) \leq (ι (n + 2) / 2)^{2}$ ; taking square roots and using step 1.1, $\sqrt{ι (k + 1)} \sqrt{ι (n - k + 1)} \leq ι (n + 2) / 2$ .

step 1.1L3L7

3.1

By the alternating series test $\sum a_{k} = \sum ε_{k} β_{k}$ converges; the same series is taken as both factors.

step 2.1step 2.2L6

3.2

Hence for every $n$ and every $k \leq n$ , $a_{k} b_{n - k} = ε_{k} ε_{n - k} β_{k} β_{n - k} = ε_{n} β_{k} β_{n - k}$ , so $c_{n} = ε_{n} \sum_{k = 0}^{n} β_{k} β_{n - k}$ and $∣ c_{n} ∣ = \sum_{k = 0}^{n} β_{k} β_{n - k}$ , the terms being positive.

step 2.1step 1.2L1L8L9

3.3

Inverting, $β_{k} β_{n - k} \geq 2 / ι (n + 2)$ for every $k \leq n$ .

step 2.3L4

4.1

Summing the $n + 1$ terms and using monotonicity of finite sums and the sum of a constant, $∣ c_{n} ∣ \geq ι (n + 1) \cdot 2 / ι (n + 2) = 2 ι (n + 1) / ι (n + 2)$ .

step 3.2step 3.3L8

5.1

Moreover $2 ι (n + 1) = ι (2 n + 2) \geq ι (n + 2)$ , since $2 n + 2 \geq n + 2$ and $ι$ is increasing; so $∣ c_{n} ∣ \geq 1$ for every $n$ .

step 4.1L3

6.1

The sequence $(c_{n})$ therefore does not converge to $0$ : the tolerance $ε = 1$ admits no index $K$ with $∣ c_{n} - 0 ∣ < 1$ for all $n \geq K$ . Hence $\sum c_{n}$ diverges.

step 5.1L10

7.1

So both factors converge while their Cauchy product diverges, and the claim [A1] is false; what is true is [L12], which asks one factor to converge absolutely, and this witness cannot satisfy that hypothesis, since otherwise Mertens' theorem would make $\sum c_{n}$ convergent, contrary to step 6.1.

step 3.1step 6.1A1L12∎

Remarks

The lower bound is not merely nonzero: it grows to $2$ . Step 4.1 gives $∣ c_{n} ∣ \geq 2 ι (n + 1) / ι (n + 2) = 2 - 2 / ι (n + 2)$ , and that bound increases to $2$ ; so the terms of the Cauchy product do not shrink at all, and the divergence is detected by the crudest test available. What the size of $∣ c_{n} ∣$ itself tends to is not determined here and is not needed.
Where the failure comes from. In $c_{n}$ every one of the $n + 1$ products $a_{k} b_{n - k}$ carries the same sign $ε_{n}$ , so no cancellation occurs within $c_{n}$ : the alternation that makes each factor converge is exactly what aligns the terms of the product. Absolute convergence of one factor, as in Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , prevents this by making the total mass finite.
The claim becomes true under other hypotheses. If all three series $\sum a_{k}$ , $\sum b_{k}$ and $\sum c_{n}$ are assumed to converge, then the sum of the product is the product of the sums; but that theorem is proved through power series and Abel's limit theorem, which are later in the reading order. The companion examples page records the same witness from the other side.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: if some grouping of a series converges then the series itself converges

Statement

False claim: if $(n_{j})$ is strictly increasing with $n_{0} = 0$ and the series of blocks $\sum_{j} B_{j}$ , $B_{j} = \sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ , converges (Series, partial sums, convergence and the sum, divergence, and the tail series), then $\sum a_{k}$ converges.

What is true is the opposite direction, Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum: convergence of $\sum a_{k}$ implies convergence of every grouping, to the same sum. Brackets may be inserted into a convergent series; they may not be removed.

The witness is the alternating sequence itself. Let $(ε_{k})$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ), with even and odd index maps $e$ and $o$ satisfying $o_{j} = e_{j} + 1$ and $e_{j + 1} = o_{j} + 1$ , and group in pairs, $n_{j} : = e_{j}$ . Every block is $ε_{e_{j}} + ε_{o_{j}} = 1 + (- 1) = 0$ , so the grouped series is $0 + 0 + \dots$ and converges to $0$ ; but $\sum ε_{k}$ diverges, its terms having absolute value $1$ and so not tending to $0$ (If a series converges then its terms tend to $0$ ).

Facts & Assumptions

Given: The alternating sequence $(ε_{k})$ with index maps $e$ and $o$ , and the grouping $n_{j} : = e_{j}$ .

[A1]

The refuted claim: if some grouping of $\sum a_{k}$ converges then $\sum a_{k}$ converges.

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{k + 1} = - ε_{k}$ , $∣ ε_{k} ∣ = 1$ , $ε_{e_{j}} = 1$ , $ε_{o_{j}} = - 1$ ; $e_{0} = 0$ , $e_{j + 1} = e_{j} + 2$ , $o_{j} = e_{j} + 1$ ; $e$ is strictly increasing (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L2]

Finite sums: the empty sum is $0$ , $\sum_{k < m + 1} x_{k} = \sum_{k < m} x_{k} + x_{m}$ , and a sum over the range ${n_{j}, \dots, n_{j + 1} - 1}$ of two indices is the sum of the two terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

Partial sums, sums and divergence of a series (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

If $\sum x_{k}$ converges then $x_{k} \to 0$ (If a series converges then its terms tend to $0$ ).

[L5]

Grouping in the true direction: if $\sum a_{k}$ converges then every grouping converges to the same sum (Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum).

Refutation

technique · direct

1.1

The map $j \mapsto n_{j} = e_{j}$ is strictly increasing with $n_{0} = e_{0} = 0$ , and $n_{j + 1} = e_{j} + 2 = n_{j} + 2$ , so each block runs over the two indices $e_{j}$ and $o_{j} = e_{j} + 1$ .

1.2

The series $\sum_{k} ε_{k}$ diverges: $∣ ε_{k} ∣ = 1$ for every $k$ , so the tolerance $ε = 1$ admits no index $K$ with $∣ ε_{k} - 0 ∣ < 1$ for all $k \geq K$ , and $(ε_{k})$ does not converge to $0$ .

L1L4

2.1

Each block is $B_{j} = ε_{e_{j}} + ε_{o_{j}} = 1 + (- 1) = 0$ .

step 1.1L1L2

3.1

The grouped series $\sum_{j} B_{j}$ has all terms $0$ , so all its partial sums are $0$ and it converges, with sum $0$ .

step 2.1L2L3

4.1

A grouping of $\sum ε_{k}$ therefore converges while $\sum ε_{k}$ does not, so the claim [A1] is false.

step 3.1step 1.2A1

5.1

What survives is [L5]: convergence of the series implies convergence of every grouping, and the implication cannot be reversed.

step 4.1A1L5∎

Remarks

The blocks hide a cancellation the grouped series cannot see. Each block is the sum of two terms of absolute value $1$ ; grouping reports only their sum, and the information that destroys convergence lives strictly inside a block. This is why Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum never looks inside a block and why its converse is hopeless in general.
What the witness does and does not isolate. Its block lengths are all equal to $2$ , so no amount of control on the block lengths alone repairs the claim. What it does violate is $a_{k} \to 0$ , which by If a series converges then its terms tend to $0$ is necessary for convergence of $\sum a_{k}$ and is invisible to the grouped series. Whether adding that condition to the hypothesis repairs the claim is not decided here and is not needed anywhere on this page.
The same series is the classical Grandi series. Grouped as $(1 - 1) + (1 - 1) + \dots$ it appears to sum to $0$ , and grouped as $1 + (- 1 + 1) + (- 1 + 1) + \dots$ it appears to sum to $1$ . Both are groupings in the sense of Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum: the second is $n = (0, 1, 3, 5, \dots)$ , strictly increasing with $n_{0} = 0$ , whose first block is the single term $a_{0}$ . So the two values are genuinely produced by two admissible groupings, and no hypothesis of that theorem excludes either; what fails is its hypothesis on the original series, which does not converge, and that is precisely why it says nothing about the grouped sums here. The same pair appears as a counterexample on the companion examples page.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: whenever both iterated sums of a double array exist, they are equal

Statement

False claim: for every array $a : N \times N \to R$ such that every row series $\sum_{j} a_{i j}$ converges, every column series $\sum_{i} a_{i j}$ converges, and both series of those sums converge, one has

$\sum_{i = 0}^{\infty} (\sum_{j = 0}^{\infty} a_{i j}) = \sum_{j = 0}^{\infty} (\sum_{i = 0}^{\infty} a_{i j}) .$

What is true is Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, whose hypothesis is on the absolute values: each row must be absolutely summable and the row totals of absolute values must themselves be summable. Without that hypothesis both iterated sums can exist and differ.

The witness is the array

$a_{i j} : = {\begin{cases} 1 & if j = i, \\ - 1 & if j = i - 1 (that is i = j + 1), \\ 0 & otherwise. \end{cases}$

Every row and every column has at most two nonzero entries, so every row series and every column series converges. Row $0$ sums to $1$ and every later row to $0$ , giving iterated sum $1$ ; every column sums to $0$ , giving iterated sum $0$ .

Facts & Assumptions

Given: The array $a$ with $a_{i i} = 1$ for every $i$ , $a_{i + 1, i} = - 1$ for every $i$ , and $a_{i j} = 0$ for all other pairs.

[A1]

The refuted claim: whenever all the row and column series and both series of their sums converge, the two iterated sums are equal.

[L1]

Finite sums: the empty sum is $0$ and $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ ; a finite sum of zeros is $0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

A series whose partial sums are constant from some index on converges to that constant, directly from the definition of a limit (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L3]

Fubini for double series, whose hypothesis is that each $\sum_{j} ∣ a_{i j} ∣$ converges and that the series of those row totals converges (Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, Absolutely convergent and conditionally convergent series, and the general starting index).

Refutation

technique · direct

1.1

Fix $i$ . The only nonzero entries in row $i$ are $a_{i i} = 1$ and, when $i \geq 1$ , $a_{i, i - 1} = - 1$ , the latter being the entry $a_{(i - 1) + 1, i - 1}$ . Both have column index below $i + 1$ , so the partial sums $\sum_{j < Q} a_{i j}$ are constant for $Q \geq i + 1$ , every further term being $0$ .

givenL1

1.2

Fix $j$ . The only nonzero entries in column $j$ are $a_{j j} = 1$ and $a_{j + 1, j} = - 1$ , so $\sum_{i < P} a_{i j} = 0$ for $P \geq j + 2$ , and the column series converges with sum $C_{j} = 0$ .

givenL1L2

2.1

Hence every row series converges: row $0$ has $\sum_{j < Q} a_{0 j} = 1$ for $Q \geq 1$ , so $R_{0} = 1$ ; and for $i \geq 1$ , $\sum_{j < Q} a_{i j} = - 1 + 1 = 0$ for $Q \geq i + 1$ , so $R_{i} = 0$ .

step 1.1L1L2

2.2

The series $\sum_{j} C_{j}$ has all terms $0$ , so it converges with sum $0$ .

step 1.2L1L2

3.1

The series $\sum_{i} R_{i}$ has partial sums equal to $1$ from index $1$ on, so it converges with sum $1$ .

step 2.1L1L2

4.1

All four convergence requirements of the claim hold, by step 2.1, step 3.1, step 1.2 and step 2.2, while the two iterated sums are $1$ and $0$ , which are different. So the claim [A1] is false.

step 3.1step 2.2A1

5.1

The hypothesis of [L3] is what fails: the row totals of absolute values are $A_{0} = 1$ and $A_{i} = 2$ for $i \geq 1$ , so $\sum_{i} A_{i}$ has unbounded partial sums and diverges, and Fubini's theorem does not apply.

step 4.1L1L3∎

Remarks

The array is as small as such an array can be. Every row and every column has at most two nonzero entries, and every entry is $0$ , $1$ or $- 1$ ; nothing is hidden in the size of the numbers. What makes the two iterated sums differ is only that the $- 1$ in each column lies one row lower than the $+ 1$ , so the cancellation happens along columns but is deferred along rows.
Both iterated sums exist, and that is the whole difficulty. A claim of this shape is not refuted by an array for which one of the sums fails to exist; the point is that existence of both is not enough, and only an absolute hypothesis makes them agree.
The failure has the same shape as rearrangement. By Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value the common value, when the absolute hypothesis holds, is also the sum along any enumeration of $N \times N$ ; an iterated sum is one particular way of exhausting the array, and choosing a different exhaustion is exactly choosing a different order of summation. The companion examples page develops the same array.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: $\prod (1 + p_{k})$ converges whenever $p_{k} \to 0$

Statement

False claim: for every sequence $(p_{k})$ of reals with $p_{k} \to 0$ , the infinite product $\prod (1 + p_{k})$ converges (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Equivalently, in the form it is usually met: an infinite product converges as soon as its factors tend to $1$ . That the factors tend to $1$ is necessary for convergence, and it is not sufficient. What decides the matter for nonnegative $p_{k}$ is For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent: $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges.

The witness is $p_{k} : = 1 / ι (k + 1)$ , with $ι (k + 1)$ the canonical natural (Canonical naturals are positive and strictly increasing). Then $p_{k} \to 0$ , while $\sum_{k} p_{k}$ is the harmonic series $\sum_{k \geq 1} 1 / k$ , which diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ); so the product diverges, its partial products satisfying $\prod_{k < n} (1 + p_{k}) \geq 1 + \sum_{k < n} p_{k}$ and hence diverging to $+ \infty$ .

Facts & Assumptions

Given: The sequence $p_{k} : = 1 / ι (k + 1)$ , its partial sums $S_{n} = \sum_{k < n} p_{k}$ and the partial products $Π_{n} = \prod_{k < n} (1 + p_{k})$ .

[A1]

The refuted claim: if $p_{k} \to 0$ then $\prod (1 + p_{k})$ converges.

[L1]

[L2]

[L3]

For nonnegative $p_{k}$ : $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges, and $1 + S_{n} \leq Π_{n}$ for every $n$ (For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent).

[L4]

Convergence of an infinite product, and divergence when no tail has partial products with a nonzero limit (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

[L5]

For a series of nonnegative terms whose partial sums are unbounded above, those partial sums diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Divergence to $+ \infty$ and to $- \infty$ , Laws of finite sums and finite products).

Refutation

technique · direct

1.1

Each $p_{k} = 1 / ι (k + 1)$ is positive, and $(p_{k})$ converges to $0$ : given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε$ satisfies $∣ p_{k} ∣ = p_{k} \leq 1 / ι (n) < ε$ for every $k \geq n$ .

givenL1

1.2

The series $\sum_{k} p_{k} = \sum_{k} 1 / ι (k + 1)$ is the $p$ -series $\sum_{k \geq 1} 1 / k$ at $p = 1$ , which diverges.

givenL2

2.1

Since the $p_{k}$ are nonnegative and $\sum p_{k}$ diverges, $\prod (1 + p_{k})$ diverges by the criterion.

step 1.1step 1.2L3

2.2

Concretely, the partial sums $S_{n}$ of the nonnegative divergent series $\sum p_{k}$ are unbounded above, so $S_{n} \to + \infty$ ; and $Π_{n} \geq 1 + S_{n}$ , so the partial products are unbounded and no tail of the product has partial products with a nonzero limit.

step 1.2L3L4L5

3.1

So $(p_{k})$ tends to $0$ while $\prod (1 + p_{k})$ diverges, and the claim [A1] is false.

step 1.1step 2.1A1

4.1

What is true is the criterion [L3]: for nonnegative terms, convergence of the product is equivalent to convergence of $\sum p_{k}$ , a strictly stronger condition than $p_{k} \to 0$ .

step 3.1A1L3∎

Remarks

The analogy with series is exact, and that is the point. For series, $a_{k} \to 0$ is necessary and not sufficient for convergence, and the harmonic series is the standard witness (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). For products, $1 + p_{k} \to 1$ is necessary and not sufficient, and the same harmonic series is the standard witness, transported through $\prod (1 + p_{k}) \geq 1 + \sum p_{k}$ .
The failure here is unbounded growth, not oscillation. The partial products increase without limit. The other way a product can fail, with partial products tending to $0$ although the factors tend to $1$ and the series of the $p_{k}$ converges, needs the signs to alternate, and is exhibited on the companion examples page.
Nothing here needs a logarithm. The single inequality $Π_{n} \geq 1 + S_{n}$ of For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent does all the work, and it is an induction on finite products.

Absolute and Conditional Convergence; Rearrangement; Products

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Absolutely convergent and conditionally convergent series, and the general starting index

Definition

Remarks

Positive and negative parts: ak=ak+−ak−ak​=ak+​−ak−​ and ∣ak∣=ak++ak−∣ak​∣=ak+​+ak−​; a series converges absolutely iff both ∑ak+∑ak+​ and ∑ak−∑ak−​ converge, and for a conditionally convergent series both diverge to +∞+∞

Statement

Facts & Assumptions

Proof

Remarks

Abel summation by parts: with An=∑k<nakAn​=∑k<n​ak​ one has ∑k<nakbk=Anbn−1−∑k<n−1Ak+1 (bk+1−bk)∑k<n​ak​bk​=An​bn−1​−∑k<n−1​Ak+1​(bk+1​−bk​) for every n≥1n≥1

Statement

Facts & Assumptions

Proof

Remarks

Dirichlet's test: if the partial sums of ∑ak∑ak​ are bounded and (bk)(bk​) is nonincreasing with bk→0bk​→0, then ∑akbk∑ak​bk​ converges

Statement

Facts & Assumptions

Proof

Remarks

The alternating series test: if (bk)(bk​) is nonincreasing with bk→0bk​→0 then ∑k(−1)kbk∑k​(−1)kbk​ converges, the sum lies between any two consecutive partial sums, and the error after nn terms is at most bnbn​

Statement

Facts & Assumptions

Proof

Remarks

Abel's test: if ∑ak∑ak​ converges and (bk)(bk​) is monotone and bounded then ∑akbk∑ak​bk​ converges

Statement

Facts & Assumptions

Proof

Remarks

Rearrangement of a series along a bijection of NN, and unconditional convergence

Definition

Remarks

Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum

Statement

Facts & Assumptions

Proof

Remarks

The Riemann series theorem: a conditionally convergent real series has, for every c∈Rc∈R, a rearrangement with sum cc, and rearrangements diverging to +∞+∞, to −∞−∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡liminf≤limsup in R‾R

Statement

Facts & Assumptions

Proof

Remarks

For a series of real numbers, unconditional convergence and absolute convergence are the same property

Statement

Facts & Assumptions

Proof

Remarks

Grouping: if ∑ak∑ak​ converges and (nj)(nj​) is strictly increasing with n0=0n0​=0, the series of blocks ∑k=njnj+1−1ak∑k=nj​nj+1​−1​ak​ converges to the same sum

Statement

Facts & Assumptions

Proof

Remarks

The Cauchy product of two series: cn=∑k=0nakbn−kcn​=∑k=0n​ak​bn−k​

Definition

Remarks

Mertens' theorem: if ∑ak∑ak​ converges absolutely to AA and ∑bk∑bk​ converges to BB, their Cauchy product converges to ABAB

Statement

Facts & Assumptions

Proof

Remarks

If ∑ak∑ak​ and ∑bk∑bk​ both converge absolutely then their Cauchy product converges absolutely, with sum ABAB

Statement

Facts & Assumptions

Proof

Remarks

Fubini for double series: if ∑i∑j∣aij∣∑i​∑j​∣aij​∣ converges then both iterated sums and the sum along every bijection N→N×NN→N×N converge to one and the same value

Statement

Facts & Assumptions

Proof

Remarks

Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors

Definition

Remarks

Statement

Facts & Assumptions

Proof

Positive and negative parts: $a_{k} = a_{k}^{+} - a_{k}^{-}$ and $∣ a_{k} ∣ = a_{k}^{+} + a_{k}^{-}$ ; a series converges absolutely iff both $\sum a_{k}^{+}$ and $\sum a_{k}^{-}$ converge, and for a conditionally convergent series both diverge to $+ \infty$

Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$

Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges

The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$

Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges

Rearrangement of a series along a bijection of $N$ , and unconditional convergence

The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$

Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum

The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$

Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$

If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$

Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value

Base- $b$ expansions: for an integer $b \geq 2$ every $x \in [0, 1)$ is the sum of $\sum_{j \geq 0} d_{j} / b^{j + 1}$ for digits $d_{j} < b$ , and the digit sequence is unique among those that are not eventually constantly $b - 1$

The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order

FALSE: $\prod (1 + p_{k})$ converges whenever $p_{k} \to 0$