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✓ 13 results · all verified · 13 also independently AI-judged

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Absolute and Conditional Convergence; Rearrangement; Products: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

$\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$

Example

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 $s \circ o \equiv - 1$ ), written $ε_{j} = (- 1)^{j}$ , and put $b_{j} : = 1 / ι (j + 1)$ , with $ι (j + 1)$ the canonical natural, positive for every $j$ (Canonical naturals are positive and strictly increasing). The alternating harmonic series is

$\sum_{j \geq 0} \frac{(- 1)^{j}}{j + 1} = \sum_{j} ε_{j} b_{j} .$

It converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index): it converges, by the alternating series test, while its series of absolute values is the harmonic series $\sum_{k \geq 1} 1 / k$ , which diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). Writing $S$ for its sum,

$\frac{1}{2} < \frac{7}{12} \leq S \leq \frac{5}{6} < 1.$

The value of $S$ is not asserted. The classical evaluation is a logarithm and is not available at this point in the reading order; what is proved here is that $S$ exists and where it lies. See Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

This is the series that gives the whole page its content: it is the standard witness for FALSE: every convergent series converges absolutely and, through 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 source of every rearrangement example below.

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ with index maps $e$ and $o$ , the sequence $b_{j} = 1 / ι (j + 1)$ , and the partial sums $t_{n} = \sum_{j < n} ε_{j} b_{j}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{j + 1} = - ε_{j}$ , $∣ ε_{j} ∣ = 1$ ; $e_{0} = 0$ , $e_{j + 1} = e_{j} + 2$ , $o_{0} = 1$ , $o_{j + 1} = o_{j} + 2$ ; $ε_{e_{j}} = 1$ 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]

The canonical naturals are positive for $n \geq 1$ and strictly increasing; reciprocation reverses the order on the positives; 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: for $(b_{j})$ nonincreasing with $b_{j} \to 0$ , $\sum ε_{j} b_{j}$ converges with sum $L$ , and $t_{e_{j}} \leq L \leq t_{o_{j}}$ for every $j$ (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]

Partial sums: $t_{0} = 0$ and $t_{n + 1} = t_{n} + ε_{n} b_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

Absolute and conditional convergence (Absolutely convergent and conditionally convergent series, and the general starting index); limits preserve non-strict inequalities (Limits preserve non-strict inequalities).

Verification

technique · direct

1.1

Every $b_{j} = 1 / ι (j + 1)$ is positive, and $(b_{j})$ is nonincreasing, since $0 < ι (j + 1) < ι (j + 2)$ .

givenL2

1.2

By [L1], $e_{1} = 2$ , $e_{2} = 4$ , $o_{1} = 3$ ; and by [L6] together with $ε_{e_{j}} = 1$ , $ε_{o_{j}} = - 1$ , the first partial sums are $t_{1} = b_{0} = 1$ , $t_{2} = 1 - 1 / 2 = 1 / 2$ , $t_{3} = 1 / 2 + 1 / 3 = 5 / 6$ and $t_{4} = 5 / 6 - 1 / 4 = 7 / 12$ .

L1L6algebra

2.1

$(b_{j})$ converges to $0$ : given a rational $ε > 0$ , take $n \geq 1$ with $1 / ι (n) < ε$ ; for $j \geq n$ one has $ι (j + 1) \geq ι (n) > 0$ , so $b_{j} \leq 1 / ι (n) < ε$ .

step 1.1L2

2.2

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

By the alternating series test the series converges; write $S$ for its sum, and $t_{e_{j}} \leq S \leq t_{o_{j}}$ holds for every $j$ .

step 1.1step 2.1L3

4.1

Taking $j = 2$ in the lower bound and $j = 1$ in the upper bound of step 3.1 gives $7 / 12 = t_{4} = t_{e_{2}} \leq S \leq t_{o_{1}} = t_{3} = 5 / 6$ .

step 3.1step 1.2

5.1

Since $1 / 2 < 7 / 12$ and $5 / 6 < 1$ , the sum satisfies $1 / 2 < S < 1$ .

step 4.1algebra

6.1

So the series converges while its series of absolute values diverges: it converges conditionally, with sum strictly between $1 / 2$ and $1$ .

step 3.1step 5.1step 2.2L7∎

Remarks

The bracketing is exactly the error bound of the test, used twice. Any pair of an even-index and an odd-index partial sum brackets $S$ , and the further out the pair is taken the tighter the bracket becomes; $t_{4}$ and $t_{3}$ are simply the first pair whose values separate $S$ strictly from $1 / 2$ and from $1$ . Taking $t_{2} = 1 / 2$ and $t_{1} = 1$ would give only the non-strict bounds.
Conditional convergence is a statement about cancellation. The terms have absolute value $1 / (j + 1)$ and their sum without signs is infinite; the series converges only because consecutive terms nearly cancel. Everything that follows on this page, that the terms may be reordered to sum to anything at all, is a consequence of exactly that.
What the bracket does not say. It gives no rate and no closed form. Better numerical bounds come from later pairs $t_{e_{j}}, t_{o_{j}}$ and cost only arithmetic; the closed form costs the logarithm.

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

Every rearrangement of $\sum_{k \geq 0} (- 1 / 2)^{k}$ converges to $2 / 3$

Example

Let $r : = - 1 / 2$ and consider $\sum_{k \geq 0} r^{k}$ , with $r^{k}$ the integer power (Integer powers $a^{m}$ ), so that the first term is $r^{0} = 1$ . Then:

$\sum_{k = 0}^{\infty} (- \frac{1}{2})^{k} = \frac{1}{1 - (- 1 / 2)} = \frac{2}{3},$

the series converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), and every rearrangement of it along a bijection of $N$ (Rearrangement of a series along a bijection of $N$ , and unconditional convergence) converges, again to $2 / 3$ .

This is the contrast case for the whole page. The alternating harmonic series ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ ) has terms with the same alternating sign pattern, tending to $0$ just as these do, and can be rearranged to any real whatever; this series cannot be rearranged to anything but $2 / 3$ . The difference is absolute convergence and nothing else, by For a series of real numbers, unconditional convergence and absolute convergence are the same property.

Facts & Assumptions

Given: $r = - 1 / 2$ and the sequence $a_{k} : = r^{k}$ (Integer powers $a^{m}$ ).

[L1]

Geometric series: for $∣ x ∣ < 1$ the series $\sum x^{k}$ converges with sum $1 / (1 - x)$ , the series starting at $k = 0$ with first term $x^{0} = 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ 1 ∣ = 1$ , and $∣ - 1 / 2 ∣ = 1 / 2$ (Basic properties of the absolute value).

[L3]

Powers: $x^{0} = 1$ and $x^{n + 1} = x^{n} x$ (Integer powers $a^{m}$ , Laws of integer exponents).

[L4]

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

[L5]

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, Rearrangement of a series along a bijection of $N$ , and unconditional convergence, Absolutely convergent and conditionally convergent series, and the general starting index).

Verification

technique · direct

1.1

An induction gives $∣ r^{k} ∣ = ∣ r ∣^{k} = (1 / 2)^{k}$ for every $k$ : at $k = 0$ both sides are $1$ , and $∣ r^{k + 1} ∣ = ∣ r^{k} r ∣ = ∣ r^{k} ∣ ∣ r ∣ = (1 / 2)^{k} (1 / 2)$ .

L2L3L4

1.2

Since $∣ r ∣ = 1 / 2 < 1$ , the series $\sum_{k} r^{k}$ converges with sum $1 / (1 - r) = 1 / (3 / 2) = 2 / 3$ .

L1L2algebra

2.1

Since $∣ 1 / 2 ∣ = 1 / 2 < 1$ , the series $\sum_{k} ∣ r^{k} ∣ = \sum_{k} (1 / 2)^{k}$ converges, with sum $1 / (1 - 1 / 2) = 2$ ; so $\sum_{k} r^{k}$ converges absolutely.

step 1.1L1L2

3.1

By Dirichlet's rearrangement theorem, for every bijection $σ$ of $N$ the series $\sum_{k} r^{σ (k)}$ converges, with the same sum $2 / 3$ .

step 2.1L5

4.1

So the series converges absolutely with sum $2 / 3$ , and every rearrangement of it converges to $2 / 3$ .

step 1.2step 2.1step 3.1∎

Remarks

The starting index matters and is stated. The series begins at $k = 0$ , and its first term is $r^{0} = 1$ ; the same series started at $k = 1$ would sum to $2 / 3 - 1 = - 1 / 3$ . For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges makes the same point for $\sum 2^{- k}$ , whose sum is $2$ from index $0$ and $1$ from index $1$ .
Nothing is checked bijection by bijection. The point of Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum is that no property of $σ$ beyond bijectivity is used; the sum of the absolute values, here $2$ , is what bounds every partial sum of every rearrangement.
The signs are a red herring. The same conclusion holds for $\sum (1 / 2)^{k}$ , whose terms are all positive; a series of nonnegative terms that converges is absolutely convergent, so no such series has an interesting rearrangement theory. The alternating signs here are chosen only to make the comparison with the alternating harmonic series exact.

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

Taking two positive terms for each negative one rearranges the alternating harmonic series to $3 / 2$ times its sum, by the identity $T_{3 n} = S_{4 n} + \frac{1}{2} S_{2 n}$

Example

Let $a_{j} : = (- 1)^{j} / ι (j + 1)$ be the terms of the alternating harmonic series, whose sum $S$ satisfies $1 / 2 < S < 1$ ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ ), and let $S_{N} : = \sum_{j < N} a_{j}$ be its partial sums.

Rearrange it by taking two positive terms for each negative one:

$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \dots$

Formally, define $σ : N \to N$ by

$σ (3 m) = 4 m, σ (3 m + 1) = 4 m + 2, σ (3 m + 2) = 2 m + 1 (m \in N),$

which is a bijection (Injection, surjection, bijection), so that $\sum_{k} a_{σ (k)}$ is a rearrangement of the alternating harmonic series (Rearrangement of a series along a bijection of $N$ , and unconditional convergence). Writing $T_{n} : = \sum_{k < n} a_{σ (k)}$ , the identity

$T_{3 n} = S_{4 n} + \frac{1}{2} S_{2 n} (n \in N)$

holds, and consequently

$\sum_{k = 0}^{\infty} a_{σ (k)} = \frac{3}{2} S .$

The value is stated relative to $S$ , and deliberately so. Texts that already have the logarithm state this example as a multiple of $\log 2$ ; that expression is not available at this point in the reading order, and the identity above needs none (Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for). Since $1 / 2 < S < 1$ , the rearranged sum lies strictly between $3 / 4$ and $3 / 2$ , and in particular differs from $S$ : a concrete instance 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 $\overline{R}$ .

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ with index maps $e_{j} = 2 j$ and $o_{j} = 2 j + 1$ ; the terms $a_{j} = ε_{j} / ι (j + 1)$ ; the partial sums $S_{N} = \sum_{j < N} a_{j}$ of the alternating harmonic series, with sum $S$ ; and $T_{n} = \sum_{k < n} a_{σ (k)}$ .

[L1]

The alternating sequence and its index maps: $e_{0} = 0$ , $e_{j + 1} = e_{j} + 2$ , $o_{0} = 1$ , $o_{j + 1} = o_{j} + 2$ ; $N$ is the disjoint union of the ranges of $e$ and $o$ , each element occurring for exactly one index; $ε_{e_{j}} = 1$ 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]

The alternating harmonic series converges, with sum $S$ satisfying $1 / 2 < S < 1$ ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ , 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).

[L3]

[L4]

The recursion theorem and the principle of induction (The recursion theorem, The principle of mathematical induction); every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

Finite sums: $\sum_{k < 0} x_{k} = 0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , additivity, scaling and splitting (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

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

[L7]

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

[L8]

A rearrangement is the composite of the terms with a bijection of $N$ (Rearrangement of a series along a bijection of $N$ , and unconditional convergence, Injection, surjection, bijection).

Verification

technique · direct

1.1

An induction gives $e_{j} = 2 j$ and $o_{j} = 2 j + 1$ for every $j$ , from $e_{0} = 0$ , $o_{0} = 1$ and the two recursions; so by [L1] every natural number is $2 i$ for exactly one $i$ or $2 i + 1$ for exactly one $i$ .

L1L4

1.2

Every natural $n$ is $3 m + r$ for exactly one pair $(m, r)$ with $r < 3$ : for existence, the set ${m^{'} : 3 m^{'} > n}$ is nonempty, containing $n + 1$ , so it has a least element $m_{0}$ , which is not $0$ since $3 \cdot 0 = 0 \leq n$ ; put $m : = m_{0} - 1$ , so $3 m \leq n < 3 m + 3$ and $r : = n - 3 m$ satisfies $r < 3$ . For uniqueness, if $3 m + r = 3 m^{'} + r^{'}$ with $r, r^{'} < 3$ and $m < m^{'}$ , then $3 m + r < 3 m + 3 \leq 3 m^{'} \leq 3 m^{'} + r^{'}$ , a contradiction; so $m = m^{'}$ and then $r = r^{'}$ .

1.3

The maps $n \mapsto 4 n$ and $n \mapsto 2 n$ are strictly increasing, so $(S_{4 n})_{n}$ and $(S_{2 n})_{n}$ are subsequences of $(S_{N})$ and both converge to $S$ .

L2L6

2.1

Applying step 1.1 twice, every natural number is exactly one of $4 i$ , $4 i + 2$ or $2 i + 1$ , for exactly one $i$ : an even number $2 m$ is $4 i$ when $m = 2 i$ and $4 i + 2$ when $m = 2 i + 1$ , and these two cases are exclusive and exhaustive by step 1.1 applied to $m$ .

step 1.1

3.1

The map $σ$ is therefore a well-defined function on $N$ , given on the unique representation $n = 3 m + r$ by the three clauses of the statement; it may equally be produced by the recursion theorem applied to the state set $N \times {0, 1, 2}$ with the cycle $(m, 0) \mapsto (m, 1) \mapsto (m, 2) \mapsto (m + 1, 0)$ . It is a bijection: by step 1.2 the pairs $(m, r)$ with $r < 3$ correspond exactly to the naturals $n = 3 m + r$ , and by step 2.1 the three clauses send those pairs bijectively onto $N$ .

step 2.1step 1.2L4L8

4.1

By [L1] and step 1.1, $ε_{4 m} = ε_{4 m + 2} = 1$ and $ε_{2 m + 1} = - 1$ , so $a_{σ (3 m)} = 1 / ι (4 m + 1)$ , $a_{σ (3 m + 1)} = 1 / ι (4 m + 3)$ and $a_{σ (3 m + 2)} = - 1 / ι (2 m + 2)$ .

step 1.1step 3.1L1

5.1

An induction on $n$ gives $T_{3 n} = S_{4 n} + \frac{1}{2} S_{2 n}$ . At $n = 0$ all three sums are empty, hence $0$ . For the step, by step 4.1 and [L5], $T_{3 (n + 1)} - T_{3 n} = 1 / ι (4 n + 1) + 1 / ι (4 n + 3) - 1 / ι (2 n + 2)$ , while $S_{4 (n + 1)} - S_{4 n} = 1 / ι (4 n + 1) - 1 / ι (4 n + 2) + 1 / ι (4 n + 3) - 1 / ι (4 n + 4)$ and $\frac{1}{2} (S_{2 (n + 1)} - S_{2 n}) = \frac{1}{2} (1 / ι (2 n + 1) - 1 / ι (2 n + 2)) = 1 / ι (4 n + 2) - 1 / ι (4 n + 4)$ ; adding the last two gives $1 / ι (4 n + 1) + 1 / ι (4 n + 3) - 2 / ι (4 n + 4)$ , and $2 / ι (4 n + 4) = 1 / ι (2 n + 2)$ .

step 4.1L3L4L5

5.2

For $3 n \leq m \leq 3 n + 2$ the difference $T_{m} - T_{3 n}$ is a sum of at most the two positive terms $1 / ι (4 n + 1)$ and $1 / ι (4 n + 3)$ , so $0 \leq T_{m} - T_{3 n} \leq 2 / ι (4 n + 1)$ .

step 4.1L3L5

6.1

Hence $T_{3 n} \to S + \frac{1}{2} S = \frac{3}{2} S$ by step 5.1 and the algebra of limits.

step 5.1step 1.3L7

7.1

Let $ε > 0$ be rational. By step 6.1 fix $N_{1}$ with $∣ T_{3 n} - \frac{3}{2} S ∣ < ε / 2$ for $n \geq N_{1}$ , and by [L3] fix $N_{2} \geq 1$ with $2 / ι (4 N_{2} + 1) < ε / 2$ , which then holds with $n$ in place of $N_{2}$ for every $n \geq N_{2}$ ; put $N : = \max {N_{1}, N_{2}}$ .

step 6.1L3choose

8.1

Let $m \geq 3 N$ and write $m = 3 n + r$ with $r < 3$ as in step 1.2; then $3 N \leq m < 3 n + 3$ , so $N < n + 1$ and $n \geq N$ . Hence $∣ T_{m} - \frac{3}{2} S ∣ \leq ∣ T_{m} - T_{3 n} ∣ + ∣ T_{3 n} - \frac{3}{2} S ∣ < ε / 2 + ε / 2 = ε$ .

step 1.2step 5.2step 7.1

9.1

Therefore $T_{m} \to \frac{3}{2} S$ : the rearranged series $\sum_{k} a_{σ (k)}$ converges with sum $\frac{3}{2} S$ , and since $1 / 2 < S < 1$ that sum lies strictly between $3 / 4$ and $3 / 2$ , so in particular it is not $S$ .

step 3.1step 8.1L2∎

Remarks

Why the identity is the right thing to prove. It compares the rearranged partial sums with two subsequences of the original partial sums, and both subsequences converge to $S$ for free. No estimate of $S$ is needed anywhere, and no closed form for it; the whole computation is an exact identity between finite sums, checked at $n = 1$ as $1 + 1 / 3 - 1 / 2 = 5 / 6$ against $7 / 12 + 1 / 4 = 5 / 6$ .
The rearrangement is explicit, unlike the one produced by 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 $\overline{R}$ . There the bijection is built by a greedy recursion depending on the whole series; here it is given by three formulas. The price is that its sum is whatever the identity says it is, rather than a prescribed target.
The three residual index classes are handled, not waved through. The identity constrains $T_{m}$ only at multiples of $3$ ; step 5.2 and step 8.1 close the gap, using that the two intervening terms are positive and tend to $0$ . An argument stopping at step 6.1 would have proved convergence of a subsequence only.

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

An explicit greedy rearrangement of the alternating harmonic series with sum $0$ , and the same recipe for any prescribed real

Example

Let $a_{j} = (- 1)^{j} / ι (j + 1)$ be the terms of the alternating harmonic series, which converges conditionally ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ ). Fix a real $c$ . The greedy rearrangement towards $c$ is the bijection $σ$ of $N$ produced by the following rule, which is exactly the construction 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 $\overline{R}$ with the constant target $c$ :

at each step, if the running sum of the terms already used is at most $c$ , take the next unused nonnegative term of the series; otherwise take the next unused negative term.

By 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 resulting rearrangement converges, with

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

For $c = 0$ the rule produces, in order,

$1, - \frac{1}{2}, - \frac{1}{4}, - \frac{1}{6}, - \frac{1}{8}, \frac{1}{3}, - \frac{1}{10}, - \frac{1}{12}, - \frac{1}{14}, - \frac{1}{16}, \frac{1}{5}, \dots$

the running sums after the successive terms being $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{12}, - \frac{1}{24}$ , then $\frac{7}{24}$ after $\frac{1}{3}$ , and so on: one positive term followed by however many negative terms are needed to bring the running sum below $0$ again.

The same series therefore has rearrangements summing to $0$ , to $S$ itself, to $\frac{3}{2} S$ (Taking two positive terms for each negative one rearranges the alternating harmonic series to $3 / 2$ times its sum, by the identity $T_{3 n} = S_{4 n} + \frac{1}{2} S_{2 n}$ ) and to every other real number, while its terms are never changed.

Facts & Assumptions

Given: The terms $a_{j} = (- 1)^{j} / ι (j + 1)$ of the alternating harmonic series, and a real number $c$ .

[L1]

The alternating harmonic series converges and does not converge absolutely, so it converges conditionally ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ , Absolutely convergent and conditionally convergent series, and the general starting index).

[L2]

For a conditionally convergent series both part series $\sum a_{j}^{+}$ and $\sum a_{j}^{-}$ diverge to $+ \infty$ , which is what keeps both supplies of terms inexhaustible (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 $+ \infty$ ).

[L3]

The Riemann series theorem: for a conditionally convergent series and every real $c$ there is a bijection $σ$ of $N$ with $\sum a_{σ (k)}$ convergent of sum $c$ ; the bijection is the greedy one described above, built by the recursion theorem on a state carrying the two counters and the running sum, with no least crossing index selected and no choice made (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 recursion theorem, The well-ordering principle, Rearrangement of a series along a bijection of $N$ , and unconditional convergence, Series, partial sums, convergence and the sum, divergence, and the tail series).

Verification

technique · direct

1.1

The alternating harmonic series converges conditionally.

givenL1

2.1

Its nonnegative terms are $a_{2 i} = 1 / ι (2 i + 1)$ , that is $1, 1 / 3, 1 / 5, \dots$ , and its negative terms are $a_{2 i + 1} = - 1 / ι (2 i + 2)$ , that is $- 1 / 2, - 1 / 4, - 1 / 6, \dots$ ; by [L2] the sums of each family are unbounded, so neither supply is exhausted at any stage of the greedy rule.

step 1.1L2

3.1

By the Riemann series theorem applied with the constant target $c$ , the greedy rule defines a bijection $σ$ of $N$ and $\sum_{k} a_{σ (k)}$ converges with sum $c$ .

step 1.1step 2.1L3

4.1

Taking $c = 0$ gives a rearrangement of the alternating harmonic series with sum $0$ , and taking $c$ arbitrary gives one with sum $c$ ; the terms used are the same in every case.

step 3.1L3∎

Remarks

The displayed initial segment is arithmetic, not a further claim. Starting from a running sum of $0$ , which is at most $c = 0$ , the rule takes the first nonnegative term $1$ ; the running sum $1$ then exceeds $0$ , so negative terms are taken until it drops below $0$ , which the successive values $\frac{1}{2}, \frac{1}{4}, \frac{1}{12}, - \frac{1}{24}$ show happens after four of them. The reader may continue the computation indefinitely; nothing in the verification depends on it.
Why the greedy rule terminates each phase. Each phase ends because the relevant part series diverges 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 $+ \infty$ ), so however far along the supply one is, enough remains to cross the target. This is the only place conditional convergence is used, and it is the reason the example has no analogue for an absolutely convergent series.
The overshoot vanishes, which is why the sum is exactly $c$ . At each crossing the running sum differs from $c$ by at most the term just used, and the terms of a convergent series tend to $0$ ; since both supplies are consumed in order and both are exhausted, the terms used at successive crossings tend to $0$ as well.

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

The period-three pattern $1, 1, - 2$ has partial sums in ${0, 1, 2}$ , so $\sum a_{k} / (k + 1)$ converges by Dirichlet's test although the alternating series test does not apply

Example

Let $(g_{k})$ be the sequence of naturals with values in ${0, 1, 2}$ defined by the recursion $g_{0} = 0$ and $g_{k + 1} = g_{k} + 1$ for $g_{k} \in {0, 1}$ , $g_{k + 1} = 0$ for $g_{k} = 2$ (The recursion theorem), and put

$a_{k} : = {\begin{cases} 1 & if g_{k} \in {0, 1}, \\ - 2 & if g_{k} = 2, \end{cases} b_{k} : = \frac{1}{ι (k + 1)} .$

So $(a_{k})$ is the repeating pattern $1, 1, - 2, 1, 1, - 2, \dots$ Its partial sums $A_{n} = \sum_{k < n} a_{k}$ take only the values $0, 1, 2$ , hence are bounded (Lower bound, bounded below, bounded set), while $(b_{k})$ is nonincreasing with $b_{k} \to 0$ . By Dirichlet's test (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 series

$\sum_{k} \frac{a_{k}}{k + 1}$

converges. It converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index), since $∣ a_{k} ∣ \geq 1$ for every $k$ and $\sum 1 / ι (k + 1)$ is the harmonic series.

The alternating series test does not reach this example. 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 a statement about $\sum ε_{k} b_{k}$ for the alternating sequence $(ε_{k})$ , whose values strictly alternate in sign; here $a_{0} = a_{1} = 1$ , so $(a_{k})$ is not that sequence, nor any constant multiple of it, and no reading of the test applies. This is the item on the page showing that Dirichlet's test is strictly stronger than the Leibniz criterion, and an alternating witness would not show it.

Facts & Assumptions

Given: The sequence $(g_{k})$ with values in ${0, 1, 2}$ defined by the displayed recursion, the terms $a_{k} \in {1, - 2}$ read off from it, $b_{k} = 1 / ι (k + 1)$ , and the partial sums $A_{n} = \sum_{k < n} a_{k}$ .

[L1]

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

[L2]

Partial sums satisfy $A_{0} = 0$ and $A_{n + 1} = A_{n} + a_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

[L4]

Dirichlet's test: bounded partial sums of $\sum a_{k}$ and a nonincreasing $(b_{k})$ with $b_{k} \to 0$ give convergence of $\sum a_{k} b_{k}$ (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, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set, Limits and Cauchy sequences of reals).

[L5]

[L6]

Direct comparison, in its divergence form: if $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum x_{k}$ diverges then $\sum y_{k}$ diverges (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 $\sum b_{k}$ ).

[L7]

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

[L8]

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

[L9]

The alternating series test is stated for $\sum ε_{k} b_{k}$ with $(ε_{k})$ the alternating sequence, which satisfies $ε_{0} = 1$ and $ε_{k + 1} = - ε_{k}$ (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 $b_{n}$ , 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 $s \circ o \equiv - 1$ ).

Verification

technique · direct

1.1

The recursion defines $(g_{k})$ as a function $N \to {0, 1, 2}$ , the transition being a total function of the set ${0, 1, 2}$ to itself; hence $(a_{k})$ is a well-defined sequence of reals with values in ${1, - 2}$ .

givenL1

1.2

Every $b_{k} = 1 / ι (k + 1)$ is positive, $(b_{k})$ is nonincreasing since $0 < ι (k + 1) < ι (k + 2)$ , and $b_{k} \to 0$ : given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε$ gives $b_{k} \leq 1 / ι (n) < ε$ for every $k \geq n$ .

givenL3

2.1

An induction gives $A_{n} = ι (g_{n})$ for every $n$ : at $n = 0$ both sides are $0$ ; and if $A_{n} = ι (g_{n})$ then, when $g_{n} \in {0, 1}$ we have $a_{n} = 1$ and $g_{n + 1} = g_{n} + 1$ , so $A_{n + 1} = ι (g_{n}) + 1 = ι (g_{n + 1})$ , while when $g_{n} = 2$ we have $a_{n} = - 2$ and $g_{n + 1} = 0$ , so $A_{n + 1} = ι (2) - 2 = 0 = ι (g_{n + 1})$ .

step 1.1L1L2

2.2

For every $k$ , $∣ a_{k} b_{k} ∣ = ∣ a_{k} ∣ b_{k} \geq b_{k} = 1 / ι (k + 1) \geq 0$ , since $∣ a_{k} ∣$ is $1$ or $2$ .

step 1.1step 1.2L7

3.1

Hence $A_{n} \in {0, 1, 2}$ for every $n$ and $∣ A_{n} ∣ \leq 2$ : the range of the partial sums is bounded.

step 2.1L7

3.2

The series $\sum_{k} 1 / ι (k + 1)$ is the $p$ -series $\sum_{k \geq 1} 1 / k$ at $p = 1$ , which diverges; so by comparison $\sum_{k} ∣ a_{k} b_{k} ∣$ diverges.

step 2.2L5L6

4.1

By Dirichlet's test, $\sum_{k} a_{k} b_{k}$ converges.

step 3.1step 1.2L4

5.1

Therefore $\sum_{k} a_{k} b_{k}$ converges conditionally: it converges by step 4.1 and does not converge absolutely by step 3.2.

step 4.1step 3.2L8

6.1

The alternating series test does not apply to this series: it is a statement about the alternating sequence $(ε_{k})$ , for which $ε_{0} = 1$ and $ε_{1} = - 1$ , whereas here $a_{0} = a_{1} = 1$ .

step 1.1L9∎

Remarks

Dirichlet's test needs only boundedness of the partial sums. Here they cycle through $0, 1, 2$ and never converge, so $\sum a_{k}$ itself diverges; the test nevertheless applies, and that is precisely what distinguishes it from Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges, which requires $\sum a_{k}$ to converge.
Why the pattern is $1, 1, - 2$ and not $1, - 1$ . The blocks must have mean zero for the partial sums to stay bounded, and they must not alternate, or the example would be covered by the alternating series test. The smallest integer pattern with both properties has period three.
The parity object is not used, and could not be. 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 $s \circ o \equiv - 1$ supplies the period-two structure of this library; a period-three pattern needs its own recursion, which is what the state set ${0, 1, 2}$ provides. No claim is made that the two constructions are instances of a common one.

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

$\sum_{j \geq 0} (- 1)^{j} (j + 3) / (j + 1)^{2}$ converges, by Abel's test with the monotone bounded factor $(j + 3) / (j + 1)$

Example

Put

$a_{j} : = \frac{(- 1)^{j}}{ι (j + 1)}, b_{j} : = \frac{ι (j + 3)}{ι (j + 1)} (j \in N),$

so that

$a_{j} b_{j} = \frac{(- 1)^{j} ι (j + 3)}{ι (j + 1)^{2}} .$

Then $\sum_{j} a_{j}$ is the alternating harmonic series, which converges ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ ), while $(b_{j})$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and bounded, with $1 < b_{j} \leq 3$ . By Abel's test (Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges) the series

$\sum_{j \geq 0} \frac{(- 1)^{j} (j + 3)}{(j + 1)^{2}}$

converges.

Neither of the two earlier tests reaches it as directly. 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 $b_{n}$ would require the sequence $ι (j + 3) / ι (j + 1)^{2}$ to be nonincreasing, which is true but is an extra computation; 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 would require a factor tending to $0$ , and $b_{j} \to 1 \neq 0$ . Abel's test is designed for exactly this shape: a convergent series multiplied by a monotone bounded factor.

Facts & Assumptions

Given: The sequences $a_{j} = (- 1)^{j} / ι (j + 1)$ and $b_{j} = ι (j + 3) / ι (j + 1)$ .

[L1]

The alternating harmonic series $\sum_{j} a_{j}$ converges ( $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ , 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 $b_{n}$ , 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 are positive for $n \geq 1$ and strictly increasing, and $ι (m + n) = ι (m) + ι (n)$ ; reciprocation reverses the order on the positives (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L3]

Abel's test: if $\sum a_{j}$ converges and $(b_{j})$ is monotone and bounded then $\sum a_{j} b_{j}$ converges (Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, A monotone sequence converges if and only if it is bounded, Lower bound, bounded below, bounded set, Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

Integer powers: $x^{2} = x \cdot x$ (Integer powers $a^{m}$ , Laws of integer exponents).

Verification

technique · direct

1.1

For every $j$ , $ι (j + 3) = ι (j + 1) + 2$ , so $b_{j} = 1 + 2 / ι (j + 1)$ , the canonical natural $ι (j + 1)$ being positive.

givenL2

1.2

For every $j$ , $a_{j} b_{j} = \frac{(- 1)^{j}}{ι (j + 1)} \cdot \frac{ι (j + 3)}{ι (j + 1)} = \frac{(- 1)^{j} ι (j + 3)}{ι (j + 1)^{2}}$ .

givenL4

2.1

The sequence $(b_{j})$ is nonincreasing: $0 < ι (j + 1) < ι (j + 2)$ gives $2 / ι (j + 2) < 2 / ι (j + 1)$ , hence $b_{j + 1} < b_{j}$ .

step 1.1L2

3.1

The sequence $(b_{j})$ is bounded, with $1 < b_{j} \leq b_{0} = 1 + 2 = 3$ : the lower bound because $2 / ι (j + 1) > 0$ , the upper because $(b_{j})$ is nonincreasing.

step 1.1step 2.1L2

4.1

By [L1] the series $\sum_{j} a_{j}$ converges, and by step 2.1 and step 3.1 the sequence $(b_{j})$ is monotone and bounded; so by Abel's test $\sum_{j} a_{j} b_{j}$ converges.

step 2.1step 3.1L1L3

5.1

By step 1.2 that series is $\sum_{j \geq 0} (- 1)^{j} (j + 3) / (j + 1)^{2}$ , which therefore converges.

step 1.2step 4.1∎

Remarks

Boundedness alone would not do. With $a_{j} = (- 1)^{j} / \sqrt{j + 1}$ convergent and $b_{j} = (- 1)^{j}$ bounded but not monotone, $\sum a_{j} b_{j} = \sum 1 / \sqrt{j + 1}$ diverges takes a convergent series and a bounded factor that is not monotone, and the product series diverges. Here monotonicity of $(b_{j})$ is what step 2.1 supplies, and it is the hypothesis doing the work.
The limit of the factor is $1$ , not $0$ , and that is the point. Since $2 / ι (j + 1) \to 0$ , the factor tends to $1$ , so 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 does not apply to $(b_{j})$ directly. Abel's test converts this into Dirichlet's by writing $b_{j} = (b_{j} - 1) + 1$ , which is exactly what its proof does.
The series converges conditionally. Its terms in absolute value are $ι (j + 3) / ι (j + 1)^{2} \geq 1 / ι (j + 1)$ , so the series of absolute values dominates the harmonic series; nothing below needs this, and it is recorded to place the example alongside $\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$ rather than alongside Every rearrangement of $\sum_{k \geq 0} (- 1 / 2)^{k}$ converges to $2 / 3$ .

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

With $a_{j} = (- 1)^{j} / \sqrt{j + 1}$ convergent and $b_{j} = (- 1)^{j}$ bounded but not monotone, $\sum a_{j} b_{j} = \sum 1 / \sqrt{j + 1}$ diverges

Statement refuted

Refuted claim: if $\sum a_{j}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $(b_{j})$ is bounded (Lower bound, bounded below, bounded set), then $\sum a_{j} b_{j}$ converges.

This is Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges with the word monotone deleted from its hypothesis on $(b_{j})$ . Deleting it destroys the theorem.

$a_{j} : = \frac{ε_{j}}{\sqrt{ι (j + 1)}}, b_{j} : = ε_{j} .$

Then $\sum a_{j}$ converges by the alternating series test, $(b_{j})$ is bounded with $∣ b_{j} ∣ = 1$ , and

$a_{j} b_{j} = \frac{ε_{j}^{2}}{\sqrt{ι (j + 1)}} = \frac{1}{\sqrt{ι (j + 1)}},$

so $\sum_{j} a_{j} b_{j}$ is $\sum_{k \geq 1} 1 / k^{1 / 2}$ , the $p$ -series at $p = 1 / 2$ , which diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Facts & Assumptions

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

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{j + 1} = - ε_{j}$ , $∣ ε_{j} ∣ = 1$ (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]

Square roots: every $t \geq 0$ has a unique $\sqrt{t} \geq 0$ with $(\sqrt{t})^{2} = t$ , and $\sqrt{t} = t^{1 / 2}$ in the notation of rational powers (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base).

[L3]

[L4]

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).

[L5]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; 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 $p > 1$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , and $x^{2} = ∣ x ∣^{2}$ (Basic properties of the absolute value).

[L7]

Abel's test, whose hypothesis on the second factor is that it be monotone and bounded (Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set).

Counterexample

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.

1.2

The sequence $(b_{j}) = (ε_{j})$ is bounded, $∣ b_{j} ∣ = 1$ for every $j$ .

L1L6

1.3

It is not monotone: $b_{0} = 1 > b_{1} = - 1$ , so it is not nondecreasing, and $b_{1} = - 1 < b_{2} = 1$ , so it is not nonincreasing.

2.1

Each $β_{j} = 1 / \sqrt{ι (j + 1)}$ is positive and $(β_{j})$ is nonincreasing, since $0 < ι (j + 1) < ι (j + 2)$ gives $0 < \sqrt{ι (j + 1)} < \sqrt{ι (j + 2)}$ .

step 1.1L2L3

2.2

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

step 1.1L2L3

3.1

By the alternating series test $\sum_{j} a_{j} = \sum_{j} ε_{j} β_{j}$ converges.

step 2.1step 2.2L4

3.2

For every $j$ , $a_{j} b_{j} = ε_{j}^{2} β_{j} = ∣ ε_{j} ∣^{2} β_{j} = β_{j} = 1 / \sqrt{ι (j + 1)}$ .

step 2.1L1L6

4.1

The series $\sum_{j} β_{j}$ is $\sum_{k \geq 1} 1 / \sqrt{k} = \sum_{k \geq 1} 1 / k^{1 / 2}$ , the $p$ -series at $p = 1 / 2$ ; since $1 / 2 > 1$ is false, it diverges.

step 3.2L2L5

5.1

So $\sum a_{j}$ converges and $(b_{j})$ is bounded, while $\sum a_{j} b_{j}$ diverges: the refuted claim fails, and the hypothesis of [L7] that is missing is precisely monotonicity of $(b_{j})$ .

step 3.1step 1.2step 1.3step 4.1L7∎

Remarks

The failure is not a matter of size. The factor $(b_{j})$ has absolute value exactly $1$ at every index, so it neither grows nor shrinks; what it does is cancel the alternation of $(a_{j})$ , and the alternation was the only reason $\sum a_{j}$ converged. A monotone factor cannot do that, which is the content of Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges.
The exponent $1 / 2$ is chosen so that both halves work. A faster decay, such as $1 / ι (j + 1)$ , would still give a divergent product series, while a faster one still, such as $1 / ι (j + 1)^{2}$ , would give an absolutely convergent $\sum a_{j}$ , for which no bounded factor can destroy convergence. The witness has to sit in the conditionally convergent range, and $p = 1 / 2$ does.
The same series appears again on this page. The Cauchy product of $\sum_{k \geq 0} (- 1)^{k} / \sqrt{k + 1}$ with itself has $∣ c_{n} ∣ \geq 1$ for every $n$ , so it diverges uses $\sum ε_{k} / \sqrt{ι (k + 1)}$ as both factors of a Cauchy product, for the same underlying reason: its terms are just barely small enough to converge with signs and not without.

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

For $∣ r ∣ < 1$ the Cauchy product of $\sum r^{k}$ with itself is $\sum (k + 1) r^{k}$ , with sum $1 / (1 - r)^{2}$

Example

Let $r \in R$ with $∣ r ∣ < 1$ and take $a_{k} = b_{k} = r^{k}$ (Integer powers $a^{m}$ ). Their Cauchy product (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ) is

$c_{n} = \sum_{k = 0}^{n} r^{k} r^{n - k} = \sum_{k = 0}^{n} r^{n} = ι (n + 1) r^{n},$

the sum of $n + 1$ copies of the same number. Both factors converge absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), so by If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ the product series converges absolutely, with

$\sum_{n = 0}^{\infty} (n + 1) r^{n} = (\sum_{k = 0}^{\infty} r^{k})^{2} = \frac{1}{(1 - r)^{2}} .$

This is the cheapest way to sum $\sum (n + 1) r^{n}$ available at this point in the reading order: no differentiation of a power series is needed, only the geometric series and Mertens' theorem.

Facts & Assumptions

Given: A real $r$ with $∣ r ∣ < 1$ , the sequences $a_{k} = b_{k} = r^{k}$ , and their Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[L1]

Geometric series: for $∣ x ∣ < 1$ , $\sum x^{k}$ converges with sum $1 / (1 - x)$ , the first term being $x^{0} = 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Powers: $x^{0} = 1$ , $x^{n + 1} = x^{n} x$ , and $x^{m + n} = x^{m} x^{n}$ (Integer powers $a^{m}$ , Laws of integer exponents).

[L3]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ 1 ∣ = 1$ (Basic properties of the absolute value).

[L4]

Finite sums: $\sum_{k < n} λ = ι (n) λ$ for a constant $λ$ , 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, Canonical naturals are positive and strictly increasing).

[L5]

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

[L6]

If both factors converge absolutely, the Cauchy product converges absolutely with sum the product of the sums (If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ , Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , Absolutely convergent and conditionally convergent series, and the general starting index).

Verification

technique · direct

1.1

An induction gives $∣ r^{k} ∣ = ∣ r ∣^{k}$ for every $k$ : at $k = 0$ both sides are $1$ , and $∣ r^{k + 1} ∣ = ∣ r^{k} r ∣ = ∣ r^{k} ∣ ∣ r ∣ = ∣ r ∣^{k} ∣ r ∣ = ∣ r ∣^{k + 1}$ .

L2L3L5

1.2

For $k \leq n$ , $r^{k} r^{n - k} = r^{k + (n - k)} = r^{n}$ , so $c_{n} = \sum_{k = 0}^{n} r^{n} = \sum_{k < n + 1} r^{n} = ι (n + 1) r^{n}$ , a sum of $n + 1$ copies of the constant $r^{n}$ .

L2L4

2.1

Since $∣ r ∣ < 1$ , both $\sum_{k} r^{k}$ and $\sum_{k} ∣ r^{k} ∣ = \sum_{k} ∣ r ∣^{k}$ converge, the first with sum $1 / (1 - r)$ ; so $\sum_{k} r^{k}$ converges absolutely.

step 1.1L1

3.1

Both factors of the Cauchy product converge absolutely, so $\sum_{n} c_{n}$ converges absolutely with sum $(1 / (1 - r))^{2} = 1 / (1 - r)^{2}$ .

step 2.1L6

4.1

By step 1.2 that series is $\sum_{n \geq 0} (n + 1) r^{n}$ , so it converges absolutely with sum $1 / (1 - r)^{2}$ .

step 1.2step 3.1∎

Remarks

The coefficient $n + 1$ counts the antidiagonal. There are exactly $n + 1$ pairs $(k, n - k)$ with $k \leq n$ , and every one of them contributes the same product $r^{n}$ ; that is the whole computation of step 1.2, and it is why the answer is a count rather than a new expression.
Absolute convergence is available for free here. The terms are powers of a fixed $r$ , so $\sum ∣ r^{k} ∣$ is again a geometric series. That is what lets If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ apply rather than only 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 it gives absolute convergence of the product as well as its value.
Compare the failing case. In The Cauchy product of $\sum_{k \geq 0} (- 1)^{k} / \sqrt{k + 1}$ with itself has $∣ c_{n} ∣ \geq 1$ for every $n$ , so it diverges the two factors again coincide, but neither converges absolutely, and the antidiagonal of $c_{n}$ has $n + 1$ terms all of the same sign and not small enough; there the count $n + 1$ is what destroys convergence rather than what produces a clean answer.

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The Cauchy product of $\sum_{k \geq 0} (- 1)^{k} / \sqrt{k + 1}$ with itself has $∣ c_{n} ∣ \geq 1$ for every $n$ , so it diverges

Statement refuted

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

The witness is a single conditionally convergent 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

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

so that $\sum a_{k}$ converges by the alternating series test. Then, as FALSE: the Cauchy product of two convergent series converges establishes,

$∣ c_{n} ∣ = \sum_{k = 0}^{n} \frac{1}{\sqrt{ι (k + 1) ι (n - k + 1)}} \geq \frac{2 ι (n + 1)}{ι (n + 2)} \geq 1 (n \in N),$

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

What this counterexample adds to the false statement is the sharp form of the bound: the lower bound $2 ι (n + 1) / ι (n + 2) = 2 - 2 / ι (n + 2)$ increases to $2$ , so $∣ c_{n} ∣$ eventually exceeds every real below $2$ . The terms of the product series therefore do not merely fail to tend to $0$ ; they stay bounded away from it by an amount approaching $2$ . Nothing here determines the asymptotic size of $∣ c_{n} ∣$ itself, only this lower bound for it; the divergence is as far from marginal as the bound makes it.

Facts & Assumptions

Given: The alternating sequence $(ε_{k})$ , the sequence $β_{k} : = 1 / \sqrt{ι (k + 1)}$ , the series $\sum a_{k}$ with $a_{k} = b_{k} = ε_{k} β_{k}$ , and its Cauchy product $(c_{n})$ .

[L1]

The series $\sum a_{k}$ converges, $∣ c_{n} ∣ = \sum_{k = 0}^{n} β_{k} β_{n - k}$ , and $∣ c_{n} ∣ \geq 2 ι (n + 1) / ι (n + 2) \geq 1$ for every $n$ ; hence $\sum c_{n}$ diverges (FALSE: the Cauchy product of two convergent series 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 $b_{n}$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, If a series converges then its terms tend to $0$ , The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L3]

The canonical naturals are positive for $n \geq 1$ and strictly increasing, with $ι (m + n) = ι (m) + ι (n)$ ; reciprocation reverses the order on the positives (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L4]

Finite sums are monotone in their terms, and the sum of $n + 1$ copies of a constant $λ$ is $ι (n + 1) λ$ (Laws of finite sums and finite products).

[L6]

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

[L7]

Convergence to $0$ of a sequence (Limits and Cauchy sequences of reals).

Counterexample

technique · direct

1.1

The series $\sum_{k} a_{k}$ converges, and its Cauchy product with itself satisfies $∣ c_{n} ∣ = \sum_{k = 0}^{n} β_{k} β_{n - k} \geq 2 ι (n + 1) / ι (n + 2) \geq 1$ for every $n$ .

givenL1

2.1

Hence $(c_{n})$ does not converge to $0$ : the tolerance $1$ admits no index $K$ with $∣ c_{n} - 0 ∣ < 1$ for all $n \geq K$ . So $\sum c_{n}$ diverges, and two convergent series can have a divergent Cauchy product.

step 1.1L1L7

2.2

The lower bound is itself informative: $2 ι (n + 1) / ι (n + 2) = 2 - 2 / ι (n + 2)$ , a quantity strictly increasing in $n$ that exceeds every real below $2$ from some index on. So $∣ c_{n} ∣ \geq 2 - 2 / ι (n + 2)$ for every $n$ , and the terms of the product series stay bounded away from $0$ by an amount approaching $2$ ; nothing here claims a value for $∣ c_{n} ∣$ itself, only this bound for it.

step 1.1L3L4

3.1

Neither factor converges absolutely, and that is exactly what the hypothesis of Mertens' theorem asks for: were $\sum ∣ a_{k} ∣$ convergent, [L6] would make $\sum c_{n}$ convergent, contradicting step 2.1.

step 2.1L6

4.1

So the refuted claim fails for this pair, and the hypothesis that repairs it is absolute convergence of one factor.

step 2.1step 2.2step 3.1L6∎

Remarks

Why the product cannot cancel. By 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 $s \circ o \equiv - 1$ the sign of $a_{k} b_{n - k}$ is $ε_{k} ε_{n - k} = ε_{n}$ , the same for every $k \leq n$ : the antidiagonal of the product array is sign-constant. So all $n + 1$ terms of $c_{n}$ add, and the AM-GM bound (The arithmetic mean, geometric mean inequality, with square roots as in Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ) shows each is at least $2 / ι (n + 2)$ .
Compare the geometric case. In For $∣ r ∣ < 1$ the Cauchy product of $\sum r^{k}$ with itself is $\sum (k + 1) r^{k}$ , with sum $1 / (1 - r)^{2}$ the antidiagonal also has $n + 1$ equal terms, but they are $r^{n}$ with $∣ r ∣ < 1$ , so the count $n + 1$ is beaten by the decay. Here the terms of the antidiagonal are at least $2 / ι (n + 2)$ each, and there are $n + 1$ of them, so the count wins.
This is the same series as in With $a_{j} = (- 1)^{j} / \sqrt{j + 1}$ convergent and $b_{j} = (- 1)^{j}$ bounded but not monotone, $\sum a_{j} b_{j} = \sum 1 / \sqrt{j + 1}$ diverges. Its terms are exactly small enough to be summable with alternating signs and not otherwise, which is what makes it the standard witness for both failures.

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The array with $a_{i i} = 1$ , $a_{i + 1, i} = - 1$ and every other entry $0$ has iterated sums $1$ and $0$

Example

Define $a : N \times N \to R$ by

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

so that the array has $1$ along the diagonal, $- 1$ immediately below it, and $0$ everywhere else. Then every row series and every column series converges, both series of those sums converge, and

$\sum_{i = 0}^{\infty} (\sum_{j = 0}^{\infty} a_{i j}) = 1, \sum_{j = 0}^{\infty} (\sum_{i = 0}^{\infty} a_{i j}) = 0.$

The two iterated sums exist and differ, which is FALSE: whenever both iterated sums of a double array exist, they are equal. What this example adds is the reason 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 does not apply: the row totals of absolute values are $A_{0} = 1$ and $A_{i} = 2$ for every $i \geq 1$ , so $\sum_{i} A_{i}$ diverges and the hypothesis of that theorem fails at its only substantive point.

Written out, the array is

$\begin{array}{cccc} 1 & 0 & 0 & \dots \\ - 1 & 1 & 0 & \dots \\ 0 & - 1 & 1 & \dots \\ 0 & 0 & - 1 & \dots \end{array}$

Every row after the first contains one $+ 1$ and one $- 1$ and so sums to $0$ ; every column contains one $+ 1$ and one $- 1$ and so sums to $0$ . The asymmetry is that the very first row has no $- 1$ to its left, and that single missing entry is the whole difference between $1$ and $0$ .

Facts & Assumptions

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

[L1]

For this array every row series and every column series converges, with row sums $R_{0} = 1$ and $R_{i} = 0$ for $i \geq 1$ and column sums $C_{j} = 0$ for every $j$ ; the two iterated sums are $1$ and $0$ (FALSE: whenever both iterated sums of a double array exist, they are equal, Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L2]

Finite sums: the empty sum is $0$ , a finite sum of zeros is $0$ , and $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ (Finite sums and finite products, by recursion, 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 (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L4]

Fubini for double series, whose hypothesis is that each $\sum_{j} ∣ a_{i j} ∣$ converges and 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).

Verification

technique · direct

1.1

Every row and every column series converges, the row sums are $R_{0} = 1$ and $R_{i} = 0$ for $i \geq 1$ , the column sums are all $0$ , and the two iterated sums are $1$ and $0$ respectively.

givenL1

1.2

The row totals of absolute values are $A_{0} = 1$ , row $0$ having the single nonzero entry $a_{00} = 1$ , and $A_{i} = 2$ for $i \geq 1$ , row $i$ having the two nonzero entries $a_{i i} = 1$ and $a_{i, i - 1} = - 1$ ; each such row series converges, being eventually constant.

givenL2

2.1

The partial sums $\sum_{i < P} A_{i}$ equal $1 + 2 (P - 1)$ for $P \geq 1$ , hence are unbounded above, so $\sum_{i} A_{i}$ diverges.

step 1.2L2L3

3.1

Therefore the hypothesis of Fubini's theorem fails for this array, and no contradiction with [L4] arises from the two iterated sums being different.

step 1.1step 2.1L4

4.1

So the array is a genuine witness: both iterated sums exist, they are $1$ and $0$ , and the absolute hypothesis that would force them to agree is exactly what it lacks.

step 1.1step 3.1∎

Remarks

The array is a discrete telescoping, and one entry falls off the edge. Each $- 1$ sits one row below the $+ 1$ of its own column: the $+ 1$ at $(j, j)$ and the $- 1$ at $(j + 1, j)$ are both in column $j$ , so every column cancels within itself and every column total is $0$ . Along a row the same two entries of that row, $a_{i i} = + 1$ and $a_{i, i - 1} = - 1$ , also cancel — but only for $i \geq 1$ . Row $0$ has no $- 1$ , there being no column $- 1$ for it to lie in, so its total is the uncancelled $+ 1$ . That single missing entry, and nothing else, is the whole difference between the two iterated sums. Which cancellation one sees depends only on the order of summation.
Nothing here is large. Every entry is $0$ , $1$ or $- 1$ , every row and column has at most two nonzero entries, and the discrepancy between the two iterated sums is exactly $1$ . The failure of Fubini's theorem without an absolute hypothesis is not a phenomenon of large or wild arrays.
The connection with rearrangement is exact. 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, under the absolute hypothesis the common value is also the sum along any enumeration of $N \times N$ ; here no such common value exists, just as a conditionally convergent series has no order-independent sum (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 $\overline{R}$ ).

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$(1 - 1) + (1 - 1) + \dots$ converges to $0$ while $\sum_{k} (- 1)^{k}$ diverges

Statement refuted

Refuted claim: if some grouping of a series converges, so does the series (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, Series, partial sums, convergence and the sum, divergence, and the tail series).

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 index map $e_{j}$ satisfying $e_{0} = 0$ and $e_{j + 1} = e_{j} + 2$ , and group the series $\sum_{k} ε_{k}$ in consecutive pairs, $n_{j} : = e_{j}$ . Each block is

$B_{j} = ε_{e_{j}} + ε_{o_{j}} = 1 + (- 1) = 0,$

so the grouped series is $0 + 0 + 0 + \dots$ and converges to $0$ , while $\sum_{k} ε_{k}$ diverges, its terms having absolute value $1$ and so not tending to $0$ (If a series converges then its terms tend to $0$ ). This is FALSE: if some grouping of a series converges then the series itself converges exhibited.

The partial sums of $\sum ε_{k}$ are $0, 1, 0, 1, \dots$ , and the grouping picks out exactly the even-indexed ones, all equal to $0$ . That is what 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 says a grouping always does: it reads a subsequence of the partial sums. A subsequence of a divergent sequence may of course converge, which is the whole of the phenomenon.

Facts & Assumptions

Given: The alternating sequence $(ε_{k})$ with index maps $e$ and $o$ , the grouping $n_{j} = e_{j}$ , and the blocks $B_{j} = \sum_{k = n_{j}}^{n_{j + 1} - 1} ε_{k}$ .

[L1]

For this grouping every block is $0$ , the grouped series converges to $0$ , and $\sum_{k} ε_{k}$ diverges (FALSE: if some grouping of a series converges then the series itself converges, If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L2]

The alternating sequence: $ε_{0} = 1$ , $ε_{k + 1} = - ε_{k}$ , $∣ ε_{k} ∣ = 1$ , $ε_{e_{j}} = 1$ , $ε_{o_{j}} = - 1$ , $e_{0} = 0$ , $e_{j + 1} = e_{j} + 2$ , $o_{j} = e_{j} + 1$ , and $e$ 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).

[L3]

Finite sums and partial sums: the empty sum is $0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L4]

Grouping in the true direction: if $\sum a_{k}$ converges then every grouping with $n_{0} = 0$ converges to the same sum, its $m$ -th partial sum being $s_{n_{m}}$ (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).

Counterexample

technique · direct

1.1

The map $j \mapsto n_{j} = e_{j}$ is strictly increasing with $n_{0} = 0$ and $n_{j + 1} = n_{j} + 2$ , so it is a grouping in the sense of [L4] and each block covers the two indices $e_{j}$ and $o_{j}$ .

L2L4

1.2

An induction gives that the partial sums $A_{n} = \sum_{k < n} ε_{k}$ take only the values $0$ and $1$ , with $A_{e_{j}} = 0$ and $A_{o_{j}} = 1$ : $A_{0} = 0$ , and each step adds $ε_{n} \in {1, - 1}$ , alternately raising and lowering the value.

L2L3

1.3

The series $\sum_{k} ε_{k}$ diverges, since $∣ ε_{k} ∣ = 1$ for every $k$ , so $(ε_{k})$ does not converge to $0$ .

L1L2

2.1

Each block is $B_{j} = ε_{e_{j}} + ε_{o_{j}} = 1 - 1 = 0$ , so the grouped series has all terms $0$ , all partial sums $0$ , and converges with sum $0$ .

step 1.1L1L2L3

3.1

So a grouping of $\sum_{k} ε_{k}$ converges while the series itself does not; the refuted claim fails.

step 2.1step 1.3

4.1

What the true statement [L4] gives is the reverse implication, and step 1.2 shows why it cannot be reversed: the grouped partial sums are the subsequence $(A_{e_{j}})$ , constantly $0$ , of a sequence that oscillates between $0$ and $1$ .

step 1.2step 3.1L4∎

Remarks

A second bracketing gives a different answer, and it is just as legitimate. Grouping as $1 + (- 1 + 1) + (- 1 + 1) + \dots$ is the grouping $n = (0, 1, 3, 5, \dots)$ , whose blocks are ${0}$ and then the pairs ${2 j - 1, 2 j}$ for $j \geq 1$ . That $n$ is strictly increasing with $n_{0} = 0$ , so it satisfies the hypotheses 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 exactly as the pairing used above does; its grouped partial sums are $A_{n_{m}}$ , that is $0, 1, 1, 1, \dots$ , so this grouping converges to $1$ while the pairing from index $0$ converges to $0$ . Two admissible groupings of one series with two different sums is the classical paradox attached to $\sum (- 1)^{k}$ , and nothing in 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 is violated: that theorem says what the grouped sums are only when the original series converges, and here it does not.
No sum is being assigned to a divergent series here. Both statements are about ordinary convergence in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series: the grouped series converges, the original does not. Nothing on this page attaches a value to $\sum (- 1)^{k}$ .
What would have to be added. By If a series converges then its terms tend to $0$ , $a_{k} \to 0$ is necessary for the original series to converge and is invisible to the grouped series; this witness violates exactly that condition, with blocks of the constant length $2$ .

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$\prod_{j \geq 0} (1 - 1 / (j + 2))$ has partial products $1 / (n + 1)$ , which tend to $0$ , so the product does not converge in the sense used here

Example

Put $p_{j} : = 1 / ι (j + 2)$ , so $0 < p_{j} \leq 1 / 2 < 1$ , and consider

$\prod_{j \geq 0} (1 - \frac{1}{j + 2}) = \prod_{j \geq 0} \frac{ι (j + 1)}{ι (j + 2)} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \dots$

Its partial products telescope:

$\prod_{j < n} (1 - \frac{1}{ι (j + 2)}) = \frac{1}{ι (n + 1)} (n \in N),$

so they tend to $0$ . Every factor is nonzero, and yet the product does not converge in the sense of Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, because no tail of it has partial products with a nonzero limit.

This is the example the definition of a convergent infinite product is written to exclude, and Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors names it for that purpose. Were a limit of $0$ admitted, this product would "converge to $0$ " with no factor equal to $0$ , and a convergent product could no longer be divided by.

The behaviour is also exactly what 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 predicts: $\sum_{j} p_{j}$ is a tail of the harmonic series and diverges, so the partial products of $\prod (1 - p_{j})$ tend to $0$ .

Facts & Assumptions

Given: The sequence $p_{j} = 1 / ι (j + 2)$ and the partial products $Q_{n} = \prod_{j < n} (1 - p_{j})$ .

[L1]

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

[L2]

The canonical naturals are positive for $n \geq 1$ , strictly increasing, and $ι (m + n) = ι (m) + ι (n)$ ; reciprocation reverses the order on the positives; 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 principle of induction on $N$ (The principle of mathematical induction).

[L4]

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

[L5]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ , with $ι (k)^{1} = ι (k)$ ; a series converges if and only if each of its tail series converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, 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).

[L6]

For $0 \leq p_{j} < 1$ with $\sum p_{j}$ divergent, the partial products of $\prod (1 - p_{j})$ tend to $0$ (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).

Verification

technique · direct

1.1

For every $j$ , $ι (j + 2) \geq ι (2) = 2 > 0$ , so $0 < p_{j} \leq 1 / 2 < 1$ and $1 - p_{j} = ι (j + 1) / ι (j + 2) > 0$ .

givenL2

2.1

An induction gives $Q_{n} = 1 / ι (n + 1)$ for every $n$ : at $n = 0$ the empty product is $1 = 1 / ι (1)$ ; and $Q_{n + 1} = Q_{n} (1 - p_{n}) = \frac{1}{ι (n + 1)} \cdot \frac{ι (n + 1)}{ι (n + 2)} = \frac{1}{ι (n + 2)}$ .

step 1.1L1L2L3

2.2

The same conclusion follows from the general criterion: $\sum_{j} p_{j} = \sum_{j} 1 / ι (j + 2)$ is the first tail series of $\sum_{j} 1 / ι (j + 1)$ , which is the harmonic series and diverges, so $\sum_{j} p_{j}$ diverges and the partial products of $\prod (1 - p_{j})$ tend to $0$ .

step 1.1L5L6

3.1

Hence $Q_{n} \to 0$ : given a rational $ε > 0$ , an $n_{0} \geq 1$ with $1 / ι (n_{0}) < ε$ gives $0 < Q_{n} = 1 / ι (n + 1) \leq 1 / ι (n_{0}) < ε$ for every $n \geq n_{0}$ .

step 2.1L2

4.1

For every $N$ the $N$ -th tail products satisfy $\prod_{j = N}^{N + n - 1} (1 - p_{j}) = Q_{N + n} / Q_{N}$ , the finite product $Q_{N}$ being positive; so they also tend to $0$ as $n$ grows, $Q_{N}$ being a fixed nonzero real.

step 1.1step 2.1step 3.1L1

5.1

Therefore no tail of the product has partial products with a nonzero limit, and $\prod_{j} (1 - p_{j})$ does not converge, although every one of its factors is nonzero.

step 4.1L4

6.1

So the partial products are $1 / ι (n + 1)$ , they tend to $0$ , and the product diverges in the sense of Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors.

step 2.1step 5.1step 2.2∎

Remarks

The telescoping is the reason the answer is exactly $1 / (n + 1)$ . Each factor is $ι (j + 1) / ι (j + 2)$ , so consecutive numerators and denominators cancel and only the first numerator $ι (1) = 1$ and the last denominator $ι (n + 1)$ survive. Written informally, $\frac{1}{2} \cdot \frac{2}{3} \dots \frac{n}{n + 1} = \frac{1}{n + 1}$ .
Why a zero limit is excluded from the definition. If it were admitted, this product would have value $0$ although no factor is $0$ ; and then from $\prod a_{k} = 0$ one could infer nothing about the factors, whereas Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors arranges that a convergent product is $0$ exactly when some factor is. The exclusion costs this one example and buys that.
The index shift is not decorative. Written as $\prod_{n \geq 1} (1 - 1 / n)$ the same product begins with the factor $1 - 1 / 1 = 0$ ; the shift to $1 - 1 / (j + 2)$ is what keeps every factor nonzero, so that the failure is genuinely about the limit and not about a vanishing factor.

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

$\prod_{j \geq 0} (1 + (- 1)^{j} / \sqrt{j + 2})$ has partial products tending to $0$ although $\sum_{j \geq 0} (- 1)^{j} / \sqrt{j + 2}$ converges

Statement refuted

Refuted claim: if $\sum p_{k}$ converges then $\prod (1 + p_{k})$ converges (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, Series, partial sums, convergence and the sum, divergence, and the tail series).

For nonnegative $p_{k}$ this is true, and 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. For signed $p_{k}$ it is false, and the witness is

$p_{j} : = \frac{(- 1)^{j}}{\sqrt{ι (j + 2)}} (j \in N),$

with $\sqrt{}$ the nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). The series $\sum_{j} p_{j}$ converges by the alternating series test. The factors $1 + p_{j}$ are all positive, since $∣ p_{j} ∣ \leq 1 / \sqrt{2} < 1$ ; nevertheless the partial products

$Π_{m} = \prod_{j < m} (1 + \frac{(- 1)^{j}}{\sqrt{j + 2}})$

tend to $0$ , so no tail of the product has partial products with a nonzero limit and the product diverges.

The mechanism, and why no logarithm is needed. Consecutive factors are paired. With $a = ι (2 i + 2)$ and $b = ι (2 i + 3)$ , so that $b - a = 1$ ,

$(1 + \frac{1}{\sqrt{a}}) (1 - \frac{1}{\sqrt{b}}) = 1 - \frac{1}{\sqrt{a b}} (1 - \frac{1}{\sqrt{a} + \sqrt{b}}) \leq 1 - \frac{1}{ι (4 i + 6)},$

and $\sum_{i} 1 / ι (4 i + 6)$ diverges. So the even partial products are dominated by $\prod_{i < n} (1 - 1 / ι (4 i + 6))$ , which tends to $0$ by 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; the odd ones differ from them by one bounded factor.

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ with index maps $e_{i} = 2 i$ and $o_{i} = 2 i + 1$ ; the sequence $p_{j} = ε_{j} / \sqrt{ι (j + 2)}$ ; the factors $f_{j} : = 1 + p_{j}$ ; and the partial products $Π_{m} = \prod_{j < m} f_{j}$ .

[L1]

The alternating sequence: $ε_{e_{i}} = 1$ , $ε_{o_{i}} = - 1$ , $∣ ε_{j} ∣ = 1$ , $e_{0} = 0$ , $e_{i + 1} = e_{i} + 2$ , $o_{i} = e_{i} + 1$ , and $N$ is the disjoint union of the two ranges (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 $(\sqrt{t})^{2} = t$ ; $\sqrt{u v} = \sqrt{u} \sqrt{v}$ and $\sqrt{}$ is strictly increasing on the nonnegative reals; and $\sqrt{t} = t^{1 / 2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base).

[L3]

The canonical naturals are positive for $n \geq 1$ , strictly increasing, with $ι (m + n) = ι (m) + ι (n)$ and $ι (m n) = ι (m) ι (n)$ for $m, n \geq 1$ ; reciprocation reverses the order on the positives; 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 $1 / n < ε$ ).

[L4]

[L5]

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

[L6]

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

[L7]

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

[L8]

For $0 \leq q_{i} < 1$ with $\sum q_{i}$ divergent, the partial products of $\prod (1 - q_{i})$ tend to $0$ (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).

[L9]

$\sum_{k \geq 1} 1 / k^{p}$ diverges at $p = 1$ ; direct comparison in its divergence form; and $\sum c x_{k}$ diverges when $c \neq 0$ and $\sum x_{k}$ diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , 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 $\sum b_{k}$ , Convergent series add and scale termwise, Integer powers $a^{m}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L10]

The squeeze theorem (The squeeze theorem).

[L11]

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

Counterexample

technique · direct

1.1

For every $j$ , $ι (j + 2) \geq ι (2) = 2$ , so $\sqrt{ι (j + 2)} \geq \sqrt{2} > 1$ and $∣ p_{j} ∣ = 1 / \sqrt{ι (j + 2)} \leq 1 / \sqrt{2} < 1$ ; hence every factor satisfies $0 < f_{j} \leq 1 + 1 / \sqrt{2} < 2$ , and every $Π_{m}$ is positive.

givenL1L2L3L6

1.2

Fix $i$ and put $a : = ι (2 i + 2)$ , $b : = ι (2 i + 3)$ , so $b - a = 1$ and $a + b = ι (4 i + 5)$ , both positive. By [L1], $f_{2 i} = 1 + 1 / \sqrt{a}$ and $f_{2 i + 1} = 1 - 1 / \sqrt{b}$ .

L1L3

1.3

An induction gives that finite products are monotone in nonnegative factors: if $0 \leq x_{i} \leq y_{i}$ for all $i < n$ then $\prod_{i < n} x_{i} \leq \prod_{i < n} y_{i}$ , since both products are nonnegative and $\prod_{i < n + 1} x_{i} = (\prod_{i < n} x_{i}) x_{n} \leq (\prod_{i < n} y_{i}) x_{n} \leq (\prod_{i < n} y_{i}) y_{n}$ .

L6L7

2.1

The sequence $β_{j} : = 1 / \sqrt{ι (j + 2)}$ is positive, nonincreasing and converges to $0$ : monotonicity from $0 < ι (j + 2) < ι (j + 3)$ and strict increase of the square root, and convergence because, given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε^{2}$ gives $ι (j + 2) > (1 / ε)^{2}$ and so $β_{j} < ε$ for every $j \geq n$ .

step 1.1L2L3

2.2

Since $\sqrt{b} - \sqrt{a} = (b - a) / (\sqrt{a} + \sqrt{b}) = 1 / (\sqrt{a} + \sqrt{b})$ , one has $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = \frac{\sqrt{b} - \sqrt{a}}{\sqrt{a} \sqrt{b}} = \frac{1}{\sqrt{a b} (\sqrt{a} + \sqrt{b})}$ , so $P_{i} : = f_{2 i} f_{2 i + 1} = 1 + \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a b}} = 1 - \frac{1}{\sqrt{a b}} (1 - \frac{1}{\sqrt{a} + \sqrt{b}})$ .

step 1.2L2algebra

2.3

Here $\sqrt{a} \geq \sqrt{2} > 1$ and $\sqrt{b} \geq \sqrt{3} > 1$ , so $\sqrt{a} + \sqrt{b} > 2$ and $1 - 1 / (\sqrt{a} + \sqrt{b}) > 1 / 2$ ; and by [L5], $\sqrt{a b} \leq (a + b) / 2 = ι (4 i + 5) / 2$ , so $1 / \sqrt{a b} \geq 2 / ι (4 i + 5)$ .

step 1.2L2L3L5

2.4

An induction gives $Π_{2 n} = \prod_{i < n} P_{i}$ for every $n$ : at $n = 0$ both are the empty product $1$ , and $Π_{2 (n + 1)} = Π_{2 n} f_{2 n} f_{2 n + 1} = Π_{2 n} P_{n}$ .

step 1.2L6L7

3.1

By the alternating series test $\sum_{j} p_{j} = \sum_{j} ε_{j} β_{j}$ converges.

step 2.1L4

3.2

Combining, $P_{i} \leq 1 - (2 / ι (4 i + 5)) \cdot \frac{1}{2} = 1 - 1 / ι (4 i + 5) \leq 1 - q_{i}$ , where $q_{i} : = 1 / ι (4 i + 6)$ , using $ι (4 i + 5) < ι (4 i + 6)$ ; and $0 < P_{i}$ by step 1.1, while $0 < q_{i} < 1$ .

step 1.1step 2.2step 2.3L3

4.1

Hence $0 < Π_{2 n} = \prod_{i < n} P_{i} \leq \prod_{i < n} (1 - q_{i})$ for every $n$ .

step 1.1step 3.2step 2.4step 1.3

4.2

The series $\sum_{i} q_{i}$ diverges: $6 (i + 1) = 6 i + 6 \geq 4 i + 6$ , so $ι (4 i + 6) \leq 6 ι (i + 1)$ and $q_{i} \geq \frac{1}{6} \cdot \frac{1}{ι (i + 1)}$ ; the series $\sum_{i} \frac{1}{6} \cdot 1 / ι (i + 1)$ diverges, being a nonzero multiple of the harmonic series, so $\sum_{i} q_{i}$ diverges by comparison.

step 3.2L3L9

5.1

By [L8] applied to $(q_{i})$ , the partial products $\prod_{i < n} (1 - q_{i})$ tend to $0$ ; with step 4.1 and the squeeze, $Π_{2 n} \to 0$ .

step 4.1step 4.2L8L10

6.1

Also $Π_{2 n + 1} = Π_{2 n} f_{2 n}$ with $0 < f_{2 n} < 2$ , so $0 < Π_{2 n + 1} < 2 Π_{2 n}$ and $Π_{2 n + 1} \to 0$ as well.

step 1.1step 5.1L6L10

7.1

Therefore $Π_{m} \to 0$ : given a rational $ε > 0$ , choose $N$ with $Π_{2 n} < ε / 2$ for all $n \geq N$ ; then for $m \geq 2 N + 1$ , writing $m$ as $2 n$ or $2 n + 1$ according to the partition of $N$ by the two index maps, in either case $n \geq N$ and $Π_{m} \leq 2 Π_{2 n} < ε$ .

step 1.1step 5.1step 6.1L1

8.1

For every $N^{'}$ the $N^{'}$ -th tail products satisfy $\prod_{j = N^{'}}^{N^{'} + n - 1} f_{j} = Π_{N^{'} + n} / Π_{N^{'}}$ with $Π_{N^{'}} > 0$ fixed, so they tend to $0$ too; no tail has partial products with a nonzero limit, and $\prod_{j} (1 + p_{j})$ diverges.

step 1.1step 7.1L6L11

9.1

So $\sum_{j} p_{j}$ converges while $\prod_{j} (1 + p_{j})$ diverges, and the refuted claim fails; the hypothesis it is missing is a sign condition, or absolute convergence of $\sum p_{j}$ , as in 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.

step 3.1step 8.1L8L11∎

Remarks

The pairing is what replaces the logarithm. The classical argument writes $\log (1 + p_{j}) = p_{j} - p_{j}^{2} / 2 + \dots$ and observes that $\sum p_{j}$ converges while $\sum p_{j}^{2}$ diverges, so the logarithms sum to $- \infty$ . That expansion is not available at this point in the reading order. Pairing consecutive factors reproduces the same effect with one algebraic identity: the first-order terms cancel to size $1 / (\sqrt{a b} (\sqrt{a} + \sqrt{b}))$ , of order $i^{- 3 / 2}$ , while the cross term $- 1 / \sqrt{a b}$ , of order $i^{- 1}$ , survives, and its sum diverges.
Absolute convergence would settle it the other way. Here $\sum_{j} ∣ p_{j} ∣ = \sum_{j} 1 / \sqrt{ι (j + 2)}$ diverges, so claim 4 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 not apply. That claim is exactly the hypothesis under which a signed product is safe.
The refinement that decides every case is deferred. For signed $(p_{j})$ with $\sum p_{j}$ convergent, the classical criterion is convergence of $\sum p_{j}^{2}$ ; for this witness $p_{j}^{2} = 1 / ι (j + 2)$ and that series diverges, which is consistent with what is proved above. The criterion itself needs the logarithm and is recorded in Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

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

$0.999 \dots = 1$ and $0.4999 \dots = 0.5$ : the second expansion of a number is exactly an eventually-all- $(b - 1)$ digit sequence

Example

Take $b = 10$ , so $β = ι (10)$ , and read a digit sequence $(d_{j})$ as the series $\sum_{j \geq 0} ι (d_{j}) / β^{j + 1}$ (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 $b - 1$ ).

All nines. The constant sequence $d_{j} = 9$ gives, by the geometric series,

$\sum_{j \geq 0} \frac{9}{10^{j + 1}} = \frac{9}{10} \cdot \frac{1}{1 - 1 / 10} = 1,$

which is the statement usually written $0.999 \dots = 1$ . Since $1 \notin [0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), this digit sequence is not the expansion of any point of $[0, 1)$ produced by 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 $b - 1$ , and no uniqueness clause is violated.

Two expansions of one number. For $x = 1 / 2$ the construction of 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 $b - 1$ returns $d_{0} = 5$ and $d_{j} = 0$ for $j \geq 1$ : the residue $r_{0} = 1 / 2$ lies in $[5 / 10, 6 / 10)$ , and $r_{1} = 10 \cdot \frac{1}{2} - 5 = 0$ , after which every digit is $0$ . But the sequence $d_{0}^{'} = 4$ , $d_{j}^{'} = 9$ for $j \geq 1$ has

$\frac{4}{10} + \sum_{j \geq 1} \frac{9}{10^{j + 1}} = \frac{4}{10} + \frac{9}{100} \cdot \frac{1}{1 - 1 / 10} = \frac{4}{10} + \frac{1}{10} = \frac{1}{2},$

which is $0.4999 \dots = 0.5$ . So two different digit sequences have the same sum, and uniqueness in 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 $b - 1$ survives only because $(d_{j}^{'})$ is terminal, being eventually constantly $b - 1 = 9$ .

This is the whole of the nonuniqueness. The uniqueness proof shows that two distinct digit sequences with the same sum must differ by one at the first index where they differ and then be all $b - 1$ against all $0$ . So excluding the terminal sequences excludes exactly one member of each such pair, and nothing else.

Facts & Assumptions

Given: The base $b = 10$ with $β = ι (10)$ , the digit sequences $d_{j} = 9$ for all $j$ ; $e_{0} = 5$ with $e_{j} = 0$ for $j \geq 1$ ; and $d_{0}^{'} = 4$ with $d_{j}^{'} = 9$ for $j \geq 1$ .

[L1]

Geometric series: for $∣ x ∣ < 1$ , $\sum_{k} x^{k}$ converges with sum $1 / (1 - x)$ , the first term being $x^{0} = 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Powers and canonical naturals: $β^{0} = 1$ , $β^{n + 1} = β^{n} β$ , $(u v)^{n} = u^{n} v^{n}$ ; $ι$ is additive and multiplicative on positive naturals and strictly increasing (Integer powers $a^{m}$ , Laws of integer exponents, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L3]

Linearity of convergent series, and a series converges if and only if each tail series does, the sum splitting as the initial partial sum plus the tail sum (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, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

Base- $b$ expansions: for $b \geq 2$ every $x \in [0, 1)$ has exactly one non-terminal digit sequence summing to $x$ , the digits being produced by the residue recursion with $d (r)$ the unique digit with $ι (d) / β \leq r < ι (d + 1) / β$ (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 $b - 1$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Limits and Cauchy sequences of reals).

Verification

technique · direct

1.1

Since $β = ι (10) \geq 2 > 1$ , one has $0 < 1 / β < 1$ , so $\sum_{k} (1 / β)^{k}$ converges with sum $1 / (1 - 1 / β) = β / (β - 1) = 10 / 9$ .

L1L2

1.2

For $x = 1 / 2$ the residue recursion gives $d_{0} = 5$ , since $ι (5) / β = 1 / 2 \leq 1 / 2 < 6 / 10 = ι (6) / β$ , and then $r_{1} = β \cdot \frac{1}{2} - ι (5) = 5 - 5 = 0$ , whence $d_{j} = 0$ and $r_{j + 1} = 0$ for every $j \geq 1$ . This sequence is non-terminal and sums to $5 / 10 = 1 / 2$ .

L2L4

2.1

Hence $\sum_{j \geq 0} 9 / β^{j + 1} = (9 / β) \sum_{j \geq 0} (1 / β)^{j} = (9 / 10) (10 / 9) = 1$ , using linearity and $β^{j + 1} = β β^{j}$ .

step 1.1L2L3

3.1

So the all-nines digit sequence has sum $1$ , and $1 \notin [0, 1)$ ; by [L4] it is therefore not the expansion of any $x \in [0, 1)$ , and it is terminal.

step 2.1L4

3.2

The sequence $(d_{j}^{'})$ with $d_{0}^{'} = 4$ and $d_{j}^{'} = 9$ for $j \geq 1$ has sum $4 / 10 + \sum_{j \geq 1} 9 / β^{j + 1}$ , and by step 2.1 with the first term removed, $\sum_{j \geq 1} 9 / β^{j + 1} = (1 / β) \sum_{j \geq 0} 9 / β^{j + 1} = 1 / 10$ ; so its sum is $4 / 10 + 1 / 10 = 1 / 2$ .

step 2.1L2L3

4.1

Thus $(e_{j})$ and $(d_{j}^{'})$ are different digit sequences with the same sum $1 / 2$ , so a real number in $[0, 1)$ can have two base- $10$ expansions.

step 1.2step 3.2

5.1

No uniqueness claim is contradicted: $(d_{j}^{'})$ is terminal, being constantly $9 = b - 1$ from index $1$ on, and [L4] asserts uniqueness only among non-terminal sequences, of which $(e_{j})$ is the one belonging to $1 / 2$ .

step 1.2step 4.1L4∎

Remarks

" $0.999 \dots = 1$ " is an identity between a series and a number, and nothing stranger. The left side denotes the sum of $\sum_{j \geq 0} 9 / 10^{j + 1}$ , which step 2.1 computes to be $1$ by the geometric series. There is no approximation and no limiting process beyond the one already in Series, partial sums, convergence and the sum, divergence, and the tail series.
Which of the two expansions the construction returns. The residue recursion of 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 $b - 1$ always takes the largest digit $d$ with $ι (d) / β \leq r$ , so it produces the expansion ending in zeros rather than the one ending in nines. That is why the constructed sequence is automatically non-terminal, as the theorem records.
The same pair exists in every base. For $b = 2$ the two expansions of $1 / 2$ are $0.1000 \dots$ and $0.0111 \dots$ ; the argument above is the case $b = 10$ of a computation that uses only $\sum_{j \geq 0} (b - 1) / b^{j + 1} = 1$ .

Absolute and Conditional Convergence; Rearrangement; Products: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

∑j≥0(−1)j/(j+1)∑j≥0​(−1)j/(j+1) converges conditionally, with sum strictly between 1/21/2 and 11

Example

Facts & Assumptions

Verification

Remarks

Every rearrangement of ∑k≥0(−1/2)k∑k≥0​(−1/2)k converges to 2/32/3

Example

Facts & Assumptions

Verification

Remarks

Taking two positive terms for each negative one rearranges the alternating harmonic series to 3/23/2 times its sum, by the identity T3n=S4n+12S2nT3n​=S4n​+21​S2n​

Example

Facts & Assumptions

Verification

Remarks

An explicit greedy rearrangement of the alternating harmonic series with sum 00, and the same recipe for any prescribed real

Example

Facts & Assumptions

Verification

Remarks

The period-three pattern 1,1,−21,1,−2 has partial sums in {0,1,2}{0,1,2}, so ∑ak/(k+1)∑ak​/(k+1) converges by Dirichlet's test although the alternating series test does not apply

Example

Facts & Assumptions

Verification

Remarks

∑j≥0(−1)j (j+3)/(j+1)2∑j≥0​(−1)j(j+3)/(j+1)2 converges, by Abel's test with the monotone bounded factor (j+3)/(j+1)(j+3)/(j+1)

Example

Facts & Assumptions

Verification

Remarks

With aj=(−1)j/j+1aj​=(−1)j/j+1​ convergent and bj=(−1)jbj​=(−1)j bounded but not monotone, ∑ajbj=∑1/j+1∑aj​bj​=∑1/j+1​ diverges

Statement refuted

Facts & Assumptions

Counterexample

Remarks

For ∣r∣<1∣r∣<1 the Cauchy product of ∑rk∑rk with itself is ∑(k+1)rk∑(k+1)rk, with sum 1/(1−r)21/(1−r)2

Example

Facts & Assumptions

Verification

Remarks

The Cauchy product of ∑k≥0(−1)k/k+1∑k≥0​(−1)k/k+1​ with itself has ∣cn∣≥1∣cn​∣≥1 for every nn, so it diverges

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The array with aii=1aii​=1, ai+1,i=−1ai+1,i​=−1 and every other entry 00 has iterated sums 11 and 00

Example

Facts & Assumptions

Verification

Remarks

(1−1)+(1−1)+…(1−1)+(1−1)+… converges to 00 while ∑k(−1)k∑k​(−1)k diverges

Statement refuted

Facts & Assumptions

Counterexample

Remarks

∏j≥0(1−1/(j+2))∏j≥0​(1−1/(j+2)) has partial products 1/(n+1)1/(n+1), which tend to 00, so the product does not converge in the sense used here

Example

Facts & Assumptions

Verification

Remarks

∏j≥0(1+(−1)j/j+2)∏j≥0​(1+(−1)j/j+2​) has partial products tending to 00 although ∑j≥0(−1)j/j+2∑j≥0​(−1)j/j+2​ converges

Statement refuted

Facts & Assumptions

Counterexample

Remarks

0.999⋯=10.999⋯=1 and 0.4999⋯=0.50.4999⋯=0.5: the second expansion of a number is exactly an eventually-all-(b−1)(b−1) digit sequence

Example

Facts & Assumptions

Verification

Remarks

Sources

$\sum_{j \geq 0} (- 1)^{j} / (j + 1)$ converges conditionally, with sum strictly between $1 / 2$ and $1$

Every rearrangement of $\sum_{k \geq 0} (- 1 / 2)^{k}$ converges to $2 / 3$

Taking two positive terms for each negative one rearranges the alternating harmonic series to $3 / 2$ times its sum, by the identity $T_{3 n} = S_{4 n} + \frac{1}{2} S_{2 n}$

An explicit greedy rearrangement of the alternating harmonic series with sum $0$ , and the same recipe for any prescribed real

The period-three pattern $1, 1, - 2$ has partial sums in ${0, 1, 2}$ , so $\sum a_{k} / (k + 1)$ converges by Dirichlet's test although the alternating series test does not apply

$\sum_{j \geq 0} (- 1)^{j} (j + 3) / (j + 1)^{2}$ converges, by Abel's test with the monotone bounded factor $(j + 3) / (j + 1)$

With $a_{j} = (- 1)^{j} / \sqrt{j + 1}$ convergent and $b_{j} = (- 1)^{j}$ bounded but not monotone, $\sum a_{j} b_{j} = \sum 1 / \sqrt{j + 1}$ diverges

For $∣ r ∣ < 1$ the Cauchy product of $\sum r^{k}$ with itself is $\sum (k + 1) r^{k}$ , with sum $1 / (1 - r)^{2}$

The Cauchy product of $\sum_{k \geq 0} (- 1)^{k} / \sqrt{k + 1}$ with itself has $∣ c_{n} ∣ \geq 1$ for every $n$ , so it diverges

The array with $a_{i i} = 1$ , $a_{i + 1, i} = - 1$ and every other entry $0$ has iterated sums $1$ and $0$

$(1 - 1) + (1 - 1) + \dots$ converges to $0$ while $\sum_{k} (- 1)^{k}$ diverges

$\prod_{j \geq 0} (1 - 1 / (j + 2))$ has partial products $1 / (n + 1)$ , which tend to $0$ , so the product does not converge in the sense used here

$\prod_{j \geq 0} (1 + (- 1)^{j} / \sqrt{j + 2})$ has partial products tending to $0$ although $\sum_{j \geq 0} (- 1)^{j} / \sqrt{j + 2}$ converges

$0.999 \dots = 1$ and $0.4999 \dots = 0.5$ : the second expansion of a number is exactly an eventually-all- $(b - 1)$ digit sequence