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

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Series: Convergence and the Nonnegative Tests

1 · Prerequisites

2 · Summary

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

Objective. A series is not a new kind of object. It is a sequence, namely the sequence of partial sums, looked at through the terms that generate it. This page makes that reduction precise in Series, partial sums, convergence and the sum, divergence, and the tail series and then spends itself on the one question the reduction leaves open: given the terms, and without ever computing the limit, how does one decide whether the partial sums converge?

Definition, and the index convention that runs through the page. Series, partial sums, convergence and the sum, divergence, and the tail series fixes the partial sums as $s_{n} = \sum_{k < n} a_{k}$ , so that $s_{0} = 0$ is the empty sum and $s_{n + 1} = s_{n} + a_{n}$ is exactly the recursion of Finite sums and finite products, by recursion with no shift. Sequences in this library are functions on $N$ and $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), while many classical series are built from expressions undefined at $0$ : $1 / k$ , $1 / k^{p}$ , $∣ a_{k} ∣^{1 / k}$ . So Series, partial sums, convergence and the sum, divergence, and the tail series also defines the series $\sum_{k \geq m} a_{k}$ of a family from a general starting index $m$ , as the series of the shifted sequence $j \mapsto a_{j + m}$ . Every statement on this page says which starting index it uses, and the ones that must start at $1$ do.

The five general facts, true for terms of any sign. A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail says convergence depends only on the terms from any index on, with the sum splitting as $s_{N}$ plus the $N$ -th tail sum. If a series converges then its terms tend to $0$ gives the necessary condition that the terms tend to $0$ , and FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges records at once that it is not sufficient. A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ is the sharp version: convergence is equivalent to a bound on every block $a_{m + 1} + \dots + a_{n}$ with $n > m \geq N$ , and it decides convergence without naming the sum. Convergent series add and scale termwise adds that convergent series may be added and scaled, and that scaling by a nonzero constant preserves both convergence and divergence. Finally If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges proves that convergence of $\sum ∣ a_{k} ∣$ forces convergence of $\sum a_{k}$ : the Cauchy criterion never asks for a sign, so the triangle inequality carries its hypothesis from the absolute values to the terms in one line. That is the only part of the theory of absolute convergence proved here, and it is the part the root and ratio tests need.

Two computable families. For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges evaluates $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ for $∣ r ∣ < 1$ and shows divergence for $∣ r ∣ \geq 1$ ; it is the comparison object behind the root and ratio tests. $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ handles $\sum (b_{k} - b_{k + 1})$ , whose partial sums are $b_{0} - b_{n}$ , and is the mechanism behind Kummer's test and the Abel-Dini theorem.

Where the sign hypothesis enters. A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum is the pivot of the page: for nonnegative terms the partial sums are nondecreasing, so convergence is exactly boundedness above of the partial sums, and the sum is their supremum. Every test that follows is an application of it. 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 $\sum b_{k}$ compares against an arbitrary series, For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each does the same through the quotient $a_{k} / b_{k}$ , phrasing the infinite case as divergence to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ) rather than as an extended limit equation, and For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges reindexes instead of comparing, squeezing blocks of $2^{n}$ terms between $2^{n}$ copies of the first and $2^{n}$ copies of the last. Condensation settles the whole $p$ -series family at a stroke: For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ proves that $\sum_{k \geq 1} 1 / k^{p}$ converges exactly when $p > 1$ , for every rational $p > 0$ , which is every exponent this page can name (Rational powers $a^{r}$ of a positive base).

The tests that read the terms against a geometric series. Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing and Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence are stated with $lim sup$ and $lim inf$ in $\overline{R}$ , so their hypotheses always mean something (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence); the root family is written $∣ a_{k + 1} ∣^{1 / (k + 1)}$ , the shift of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ , since $∣ a_{k} ∣^{1 / k}$ is undefined at $k = 0$ . What the comparison with a geometric series delivers in each case is convergence of $\sum ∣ a_{k} ∣$ ; If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges carries it the rest of the way, so both tests are stated in their standard form, concluding convergence of $\sum a_{k}$ itself. Whenever the ratio test decides, the root test decides the same way, and the converse fails reads the chain of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ through the two tests: whenever the ratio test decides, the root test decides the same way, and the standard witness $a_{k} = 2^{- k + (- 1)^{k}}$ shows the converse fails. FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges records why the divergence half of the ratio test is stated with $lim inf$ and a strict inequality.

The tests that read the rate at which the ratios approach $1$ . Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges is a family of criteria, one for each positive weight sequence $(ζ_{k})$ , and its strength is the strength of the divergent series $\sum 1 / ζ_{k}$ it carries. Positivity of the terms is load bearing there, not a normalisation: with $ζ_{k} = 1$ the convergent series $\sum (- 1 / 2)^{k}$ satisfies every hypothesis of the divergence half, so the half is false without it. Constant weights recover the ratio test (Kummer with $ζ_{k} = 1$ recovers the ratio test) and the weights $k + 1$ give Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence, whose comparison series is the harmonic one. Gauss: for positive terms, if $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ , some constant $C$ and some rational $ε > 0$ , the series converges iff $h > 1$ covers the case Raabe leaves open: under the expansion $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ the series converges exactly when $h > 1$ , and the borderline $h = 1$ is proved divergent without any logarithm, by a telescoping product estimate that bounds $k a_{k}$ below by a constant.

The hierarchy has no last term. For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges divides the terms of a divergent series of positive terms by the running total, producing a series that still diverges, and by the square of the running total, producing one that converges. FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test draws the consequence: the first of those has eventually smaller terms than the series it came from, so no divergent series of positive terms is slowest, no comparison test is universal, and the sequence of refinements on this page cannot terminate. How the nonnegative tests are ordered by strength, and which of them this page cannot state without the logarithm collects exactly which comparisons are proved here, which are deliberately not claimed, and which classical tests, Bertrand's and the integral test among them, cannot even be stated at this point because the logarithm and the integral do not yet exist in this library.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Series, partial sums, convergence and the sum, divergence, and the tail series

Definition

Throughout, $R$ is the complete ordered field (The real numbers) and a sequence of reals is a function $a : N \to R$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), written $(a_{k})$ ; recall that $N$ contains $0$ .

Partial sums. Let $(a_{k})$ be a sequence of reals. Its sequence of partial sums is

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

the finite sum of Finite sums and finite products, by recursion. In particular $s_{0} = 0$ , the empty sum, and $s_{n + 1} = s_{n} + a_{n}$ for every $n$ , those being exactly the two recursion clauses that define the finite sum. Note that $s_{n}$ is the sum of the $n$ terms $a_{0}, \dots, a_{n - 1}$ , so the index $n$ counts terms rather than naming the last one.

Convergence, the sum, divergence. The series of $(a_{k})$ , written $\sum a_{k}$ , converges when the sequence $(s_{n})$ of partial sums converges (Limits and Cauchy sequences of reals), and then the sum of the series is

$\sum_{k = 0}^{\infty} a_{k} : = \lim_{n} s_{n} .$

The series diverges when $(s_{n})$ does not converge. A convergent sequence of reals has exactly one limit (A sequence has at most one limit), so the displayed symbol names a single real number and nothing further has to be checked for it to be well defined.

Series with a general starting index. Let $m \in N$ and let $a$ be a function on ${k \in N : k \geq m}$ , which we call a family from $m$ and write $(a_{k})_{k \geq m}$ . The series

$\sum_{k \geq m} a_{k}$

is by definition the series of the sequence $b_{j} : = a_{j + m}$ , $j \in N$ , which is a genuine sequence of reals; it converges exactly when that series converges, and its sum is then written $\sum_{k = m}^{\infty} a_{k}$ . Its partial sums are

$\sum_{k = m}^{n - 1} a_{k} = \sum_{j < n - m} a_{m + j} (n \geq m),$

in the notation of Finite sums and finite products, by recursion, the value at $n = m$ being the empty sum $0$ . A sequence on $N$ is the case $m = 0$ , and the two readings of $\sum a_{k}$ agree there, since $b_{j} = a_{j}$ .

This clause is not a convenience. Sequences in this library are functions on $N$ and $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), while many of the classical series are built from expressions that are undefined at the index $0$ : $1 / k$ , $1 / k^{p}$ and $∣ a_{k} ∣^{1 / k}$ all require $k \geq 1$ . Writing such a series as $\sum_{k \geq 1} a_{k}$ names an honest object, whereas writing it as a sequence on $N$ would require a value at an index where the defining expression has none. Every statement on this page says which starting index it uses.

Tail series. For $N \in N$ , the $N$ -th tail series of $\sum a_{k}$ is $\sum_{k \geq N} a_{k}$ , that is the series of the $N$ -th tail $a^{(N)}$ of Sequences of reals: bounded, eventually, frequently, tails, subsequences, whose terms are $a_{j}^{(N)} = a_{j + N}$ . The $0$ -th tail series is the series itself.

Remarks

"Diverges" here means "does not converge", and nothing more. A divergent series may have partial sums that run away to $+ \infty$ , or to $- \infty$ , or that oscillate without settling anywhere. The three behaviours are not distinguished by the word, and no statement on this page uses "diverges" to mean "the partial sums are unbounded" unless it says so.
The symbol $\sum_{k = 0}^{\infty} a_{k}$ is defined only for a convergent series. It denotes a real number, not a formal object, and it is illegitimate to write it down before convergence has been established. Where a proof needs to speak of the series without knowing whether it converges, it speaks of $(a_{k})$ and of $(s_{n})$ .
Two indices, doing different work. The index $k$ runs over the terms and is bound; the index $n$ runs over the partial sums and is the variable in which the limit is taken. Confusing them is the commonest slip in the subject, and it is the reason the definition above fixes $s_{n} = \sum_{k < n} a_{k}$ rather than $\sum_{k \leq n} a_{k}$ : with this choice the recursion $s_{n + 1} = s_{n} + a_{n}$ is the one supplied by Finite sums and finite products, by recursion, with no shift anywhere.

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

A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail

Statement

Let $(a_{k})$ be a sequence of reals with partial sums $s_{n} = \sum_{k < n} a_{k}$ , let $N \in N$ , and let $t_{j} : = \sum_{i < j} a_{N + i}$ be the partial sums of the $N$ -th tail series $\sum_{k \geq N} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

$t_{j} = s_{j + N} - s_{N}$ for every $j \in N$ ;
$\sum a_{k}$ converges if and only if its $N$ -th tail series converges, and in that case $\sum_{k = 0}^{\infty} a_{k} = s_{N} + \sum_{k = N}^{\infty} a_{k};$
hence the following are equivalent: $\sum a_{k}$ converges; every tail series of $\sum a_{k}$ converges; some tail series of $\sum a_{k}$ converges.

In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals, a natural number $N$ , the partial sums $s_{n} = \sum_{k < n} a_{k}$ and the partial sums $t_{j} = \sum_{i < j} a_{N + i}$ of the $N$ -th tail series (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

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

[L2]

Convergence depends only on the tail: a sequence $(x_{n})$ converges to $x$ if and only if its $N$ -th tail $(x_{j + N})_{j}$ converges to $x$ (Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

Algebra of limits: if $x_{j} \to x$ and $y_{j} \to y$ then $x_{j} + y_{j} \to x + y$ and $x_{j} - y_{j} \to x - y$ (Algebra of limits: sums, scalar multiples, products and quotients).

[L4]

A constant sequence $y_{j} = c$ converges to $c$ , immediately from the definition of a limit (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Fix $j \in N$ and put $n : = j + N$ , so that $N \leq n$ and $n - N = j$ .

given

1.2

The family $(s_{j + N})_{j \in N}$ is the $N$ -th tail of the sequence $(s_{n})$ of partial sums.

givenL2

2.1

Splitting the partial sum $s_{n}$ at $N$ gives $s_{n} = \sum_{k < N} a_{k} + \sum_{k = N}^{n - 1} a_{k} = s_{N} + \sum_{i < j} a_{N + i} = s_{N} + t_{j}$ .

step 1.1L1

2.2

Suppose $\sum a_{k}$ converges, say $s_{n} \to S$ ; then the $N$ -th tail $(s_{j + N})_{j}$ converges to $S$ .

givenstep 1.2L2

3.1

Claim 1 follows: $t_{j} = s_{j + N} - s_{N}$ for every $j \in N$ .

step 2.1algebra

4.1

The constant sequence with value $s_{N}$ converges to $s_{N}$ , so $t_{j} = s_{j + N} - s_{N} \to S - s_{N}$ , and the $N$ -th tail series converges with sum $S - s_{N}$ .

step 2.2step 3.1L3L4

4.2

Conversely, suppose the $N$ -th tail series converges, say $t_{j} \to T$ ; then $s_{j + N} = s_{N} + t_{j} \to s_{N} + T$ by the same two rules.

step 3.1L3L4

5.1

So the $N$ -th tail of $(s_{n})$ converges to $s_{N} + T$ , hence $(s_{n})$ itself converges to $s_{N} + T$ and $\sum a_{k}$ converges.

step 4.2step 1.2L2

6.1

Claim 2 follows: the two convergences are equivalent, and when they hold the sums are related by $S = s_{N} + T$ , which is the displayed identity.

step 4.1step 5.1algebra

7.1

Claim 3 follows as well: if $\sum a_{k}$ converges then by claim 2 every tail series converges; every tail series converging trivially gives some tail series converging, the family of tails being indexed by all of $N$ ; and if some tail series, say the $N$ -th, converges then by claim 2 again $\sum a_{k}$ converges.

step 6.1∎

Remarks

Where the hypothesis-free character comes from. Nothing here assumes the terms have a sign or that any series converges: claim 1 is an identity between finite sums, valid always, and claims 2 and 3 are read off from it by two standard limit rules. That is why the lemma may be used to move between a series and its tails inside proofs that have not yet decided the convergence question.
The sum does change. Only the fact of convergence is tail invariant. The identity in claim 2 is the exact bookkeeping: discarding the first $N$ terms lowers the sum by $s_{N}$ , and no more.

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

If a series converges then its terms tend to $0$

Statement

Let $(a_{k})$ be a sequence of reals. If the series $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the sequence $(a_{k})$ converges (Limits and Cauchy sequences of reals), with

$\lim_{k} a_{k} = 0.$

Equivalently, in the contrapositive form in which the lemma is almost always used: if $(a_{k})$ does not converge to $0$ then $\sum a_{k}$ diverges.

The same statement holds for a series $\sum_{k \geq m} a_{k}$ with a general starting index, with the conclusion that $(a_{j + m})_{j \in N}$ converges to $0$ : that series is by definition the series of the sequence $b_{j} = a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), so it is the statement above applied to $(b_{j})$ .

The converse is false; it is refuted by FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals whose series converges, with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[A1]

$(s_{n})$ converges to some real $S$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L1]

The recursion clause of the finite sum: $s_{n + 1} = s_{n} + a_{n}$ for every $n \in N$ (Finite sums and finite products, by recursion, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Convergence depends only on the tail: if $(x_{n})$ converges to $x$ then so does its $1$ -st tail $(x_{n + 1})_{n}$ (Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

Algebra of limits: if $x_{n} \to x$ and $y_{n} \to y$ then $x_{n} - y_{n} \to x - y$ (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct

1.1

The sequence $(s_{n + 1})_{n \in N}$ is the $1$ -st tail of $(s_{n})$ , and it converges to $S$ .

A1L2

1.2

From the recursion clause, $a_{n} = s_{n + 1} - s_{n}$ for every $n \in N$ .

L1algebra

2.1

Hence $(a_{n})$ is the difference of two sequences converging to $S$ , so it converges to $S - S = 0$ .

step 1.1step 1.2A1L3

3.1

That is the claim, and its contrapositive is the assertion that a series whose terms do not tend to $0$ diverges.

step 2.1∎

Remarks

This is a necessary condition and never a sufficient one. It rules a series out; it never rules one in. The harmonic series has terms tending to $0$ and diverges, which is exactly the content of FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges.
What the proof actually uses. Only that the partial sums converge and that consecutive partial sums differ by a term. No sign hypothesis is placed on $(a_{k})$ , and none is available at this point on the page.

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

A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$

Statement

Let $(a_{k})$ be a sequence of reals, with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then $\sum a_{k}$ converges if and only if

$for every real ε > 0 there is N \in N such that ∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε for all n > m \geq N .$

The block $\sum_{k = m + 1}^{n} a_{k}$ is the finite sum $a_{m + 1} + \dots + a_{n}$ of Finite sums and finite products, by recursion, and it equals $s_{n + 1} - s_{m + 1}$ .

This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

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

[L2]

Every convergent sequence of reals is Cauchy (Every convergent sequence is Cauchy).

[L3]

Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

[L4]

Cauchyness reads: for every rational $ε > 0$ there is $K$ with $∣ s_{p} - s_{q} ∣ < ε$ for all $p, q \geq K$ (Limits and Cauchy sequences of reals). Rational and real tolerances give the same condition, because every real $ε > 0$ exceeds some rational $1 / n$ with $n \geq 1$ natural (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ), and conversely every positive rational is a positive real.

Proof

technique · direct

1.1

For all naturals $m < n$ , splitting $s_{n + 1}$ at the index $m + 1 \leq n + 1$ gives $s_{n + 1} = s_{m + 1} + \sum_{k = m + 1}^{n} a_{k}$ , hence $\sum_{k = m + 1}^{n} a_{k} = s_{n + 1} - s_{m + 1}$ .

L1algebra

1.2

Conversely, every pair $p > q \geq 1$ of naturals is of the form $p = n + 1$ , $q = m + 1$ with $n = p - 1 > m = q - 1 \geq 0$ .

givenalgebra

1.3

Suppose $\sum a_{k}$ converges, that is $(s_{n})$ converges; then $(s_{n})$ is Cauchy.

givenL2

1.4

Suppose conversely that the stated condition holds, and let a rational $ε > 0$ be given; take $N$ for $ε$ as in the condition and put $K : = N + 1$ .

givenchoose

2.1

Let a real $ε > 0$ be given, choose a rational $ε^{'}$ with $0 < ε^{'} < ε$ , and take $K$ for $ε^{'}$ as in the Cauchy condition; put $N : = K$ .

step 1.3L4choose

2.2

Let $p, q \geq K$ . If $p = q$ then $∣ s_{p} - s_{q} ∣ = 0 < ε$ ; otherwise one of them exceeds the other, and by symmetry we may take $p > q \geq K \geq 1$ .

step 1.4algebra

3.1

For all $n > m \geq N$ one has $n + 1 > m + 1 \geq K$ , so $∣ \sum_{k = m + 1}^{n} a_{k} ∣ = ∣ s_{n + 1} - s_{m + 1} ∣ < ε^{'} < ε$ , which is the stated condition.

step 2.1step 1.1L4

3.2

Writing $p = n + 1$ and $q = m + 1$ gives $n > m$ and $m = q - 1 \geq K - 1 = N$ , so the condition applies and $∣ s_{p} - s_{q} ∣ = ∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε$ .

step 2.2step 1.2step 1.1step 1.4

4.1

So $(s_{n})$ is Cauchy, hence converges, hence $\sum a_{k}$ converges.

step 2.2step 3.2L4L3

5.1

The two implications together are the stated equivalence.

step 3.1step 4.1∎

Remarks

The criterion is stated over blocks, not over partial sums, on purpose. In applications one estimates a run of consecutive terms $a_{m + 1}, \dots, a_{n}$ directly; the translation into $∣ s_{n + 1} - s_{m + 1} ∣$ is step 1.1 and is done once here so that no later proof has to repeat it.
Taking $n = m + 1$ recovers the term test. The single-term block gives $∣ a_{m + 1} ∣ < ε$ for all $m \geq N$ , which is $a_{k} \to 0$ ; so If a series converges then its terms tend to $0$ is the weakest consequence of this criterion. The criterion is strictly stronger, since it constrains arbitrarily long blocks and not only single terms.

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

Convergent series add and scale termwise

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals whose series converge (Series, partial sums, convergence and the sum, divergence, and the tail series), and let $c \in R$ . Then:

$\sum (a_{k} + b_{k})$ converges, with $\sum_{k = 0}^{\infty} (a_{k} + b_{k}) = \sum_{k = 0}^{\infty} a_{k} + \sum_{k = 0}^{\infty} b_{k}$ ;
$\sum c a_{k}$ converges, with $\sum_{k = 0}^{\infty} c a_{k} = c \sum_{k = 0}^{\infty} a_{k}$ .

Moreover, for $c \neq 0$ and an arbitrary sequence $(a_{k})$ , whose series is not assumed to converge:

$\sum c a_{k}$ converges if and only if $\sum a_{k}$ converges. Equivalently, $\sum a_{k}$ diverges if and only if $\sum c a_{k}$ diverges.

Claim 3 is the form used whenever a comparison is made against a constant multiple of a known series.

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals and $c \in R$ , with partial sums $s_{n} = \sum_{k < n} a_{k}$ and $t_{n} = \sum_{k < n} b_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Additivity and scaling of finite sums: $\sum_{k < n} (a_{k} + b_{k}) = \sum_{k < n} a_{k} + \sum_{k < n} b_{k}$ and $\sum_{k < n} c a_{k} = c \sum_{k < n} a_{k}$ (Laws of finite sums and finite products).

[L2]

Algebra of limits: if $x_{n} \to x$ and $y_{n} \to y$ then $x_{n} + y_{n} \to x + y$ and $c x_{n} \to c x$ (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct

1.1

The partial sums of $\sum (a_{k} + b_{k})$ are $\sum_{k < n} (a_{k} + b_{k}) = s_{n} + t_{n}$ , and those of $\sum c a_{k}$ are $\sum_{k < n} c a_{k} = c s_{n}$ .

givenL1

1.2

Assume $\sum a_{k}$ and $\sum b_{k}$ converge, say $s_{n} \to S$ and $t_{n} \to T$ .

given

2.1

Then $s_{n} + t_{n} \to S + T$ , so $\sum (a_{k} + b_{k})$ converges with sum $S + T$ , which is claim 1.

step 1.2step 1.1L2

2.2

Likewise $c s_{n} \to c S$ , so $\sum c a_{k}$ converges with sum $c S$ , which is claim 2.

step 1.2step 1.1L2

3.1

For claim 3, let $c \neq 0$ and let $(a_{k})$ be arbitrary. If $\sum a_{k}$ converges then $\sum c a_{k}$ converges by claim 2.

step 2.2

3.2

Conversely, if $\sum c a_{k}$ converges then applying claim 2 to the sequence $(c a_{k})$ and the scalar $1 / c$ , which exists since $c \neq 0$ , shows that $\sum (1 / c) (c a_{k}) = \sum a_{k}$ converges.

step 2.2algebra

4.1

The two implications are claim 3, and its contrapositive form is the statement about divergence.

step 3.1step 3.2∎

Remarks

There is no product rule here, and there is no rule for $\sum a_{k} b_{k}$ . The proof works because a finite sum is additive and homogeneous, and neither property has an analogue for products. Multiplying series is a genuinely harder question, requiring absolute convergence, and it is not treated on this page.
Claim 3 needs $c \neq 0$ and nothing else. In particular it does not need either series to converge, which is what makes it usable in the divergence direction: scaling a divergent series by a nonzero constant leaves it divergent.

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

If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges

Statement

Let $(a_{k})$ be a sequence of reals. If the series $\sum ∣ a_{k} ∣$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the series $\sum a_{k}$ converges.

A series with the property that $\sum ∣ a_{k} ∣$ converges is called absolutely convergent; the lemma says that absolute convergence implies convergence.

The same statement holds for a family from a general starting index $m$ , being this statement applied to the shifted sequence $j \mapsto a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

The converse is false, and the standard witness is the alternating harmonic series. That witness is not available on this page: its convergence is the alternating series test, which is not proved here. Nothing below asserts a converse, and no item on this page uses one.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals such that the series $\sum ∣ a_{k} ∣$ converges, with partial sums as in Series, partial sums, convergence and the sum, divergence, and the tail series and finite sums as in Finite sums and finite products, by recursion.

[L1]

The Cauchy criterion for series: $\sum d_{k}$ converges if and only if for every real $ε > 0$ there is $N \in N$ with $∣ \sum_{k = p + 1}^{n} d_{k} ∣ < ε$ for all $n > p \geq N$ (A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Triangle inequality for finite sums: $∣ \sum_{k < d} u_{k} ∣ \leq \sum_{k < d} ∣ u_{k} ∣$ (Triangle inequality for finite sums); the block $\sum_{k = p + 1}^{n} u_{k}$ is by definition the finite sum $\sum_{t < n - p} u_{p + 1 + t}$ (Finite sums and finite products, by recursion), so applying the inequality to the shifted sequence $t \mapsto u_{p + 1 + t}$ gives $∣ \sum_{k = p + 1}^{n} u_{k} ∣ \leq \sum_{k = p + 1}^{n} ∣ u_{k} ∣$ for all naturals $n > p$ .

[L3]

Monotonicity of finite sums: if $x_{t} \geq 0$ for all $t < d$ then $\sum_{t < d} x_{t} \geq 0$ (Laws of finite sums and finite products).

[L4]

Absolute value: $∣ u ∣ \geq 0$ for every real $u$ , and $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value).

[L5]

Convergence of a real sequence, and the fact that the real and rational formulations of a tolerance agree (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Let $ε > 0$ be an arbitrary real; since $\sum ∣ a_{k} ∣$ converges, the Cauchy criterion applied to the sequence $(∣ a_{k} ∣)$ supplies $N \in N$ with $∣ \sum_{k = p + 1}^{n} ∣ a_{k} ∣ ∣ < ε$ for all $n > p \geq N$ .

L1L5choose

1.2

For all naturals $n > p$ the block $\sum_{k = p + 1}^{n} ∣ a_{k} ∣$ is a finite sum of nonnegative terms, hence nonnegative, hence equal to its own absolute value.

L2L3L4

2.1

So for all $n > p \geq N$ one has $∣ \sum_{k = p + 1}^{n} a_{k} ∣ \leq \sum_{k = p + 1}^{n} ∣ a_{k} ∣ = ∣ \sum_{k = p + 1}^{n} ∣ a_{k} ∣ ∣ < ε$ .

step 1.1step 1.2L2

3.1

As $ε > 0$ was arbitrary, the sequence $(a_{k})$ satisfies the Cauchy criterion, so $\sum a_{k}$ converges.

step 2.1L1∎

Remarks

Nothing here identifies the two sums, and they are in general different. What is proved is that the second series converges, not that it converges to the same value; the bound $∣ \sum_{k = 0}^{\infty} a_{k} ∣ \leq \sum_{k = 0}^{\infty} ∣ a_{k} ∣$ is true and is not needed anywhere on this page, so it is not proved here.
Why the Cauchy criterion is the right instrument. The terms have no sign, so A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum does not apply to $\sum a_{k}$ and boundedness of its partial sums would prove nothing. The Cauchy criterion is the one convergence test on this page that never names a candidate sum and never asks for a sign, and the whole proof is the observation that its hypothesis for $(∣ a_{k} ∣)$ implies its hypothesis for $(a_{k})$ , term by term, through one application of the finite triangle inequality.
What this unlocks on this page. The root test (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing) and the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence) each produce convergence of $\sum ∣ a_{k} ∣$ directly, by comparison with a geometric series; with this lemma both reach their standard conclusion, the convergence of $\sum a_{k}$ itself. Without it their convergence halves would be strictly weaker than the classical statements.
The systematic theory is elsewhere. Rearrangement, the Riemann series theorem, conditional convergence and products of series all belong with absolute convergence and are developed on a later page of this track. This lemma is only the one implication those two tests need.

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

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges

Statement

Let $r \in R$ and let $r^{k}$ be the integer power (Integer powers $a^{m}$ ), so that $r^{0} = 1$ for every $r$ , including $r = 0$ .

If $∣ r ∣ < 1$ then the series $\sum r^{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $\sum_{k = 0}^{\infty} r^{k} = \frac{1}{1 - r} .$
If $∣ r ∣ \geq 1$ then $\sum r^{k}$ diverges.

The series starts at $k = 0$ and its first term is $r^{0} = 1$ ; in particular $\sum_{k = 0}^{\infty} 2^{- k} = 2$ , while the series starting at $k = 1$ sums to $1$ . Which starting index is meant has to be said, and it is said here.

Facts & Assumptions

Given: A real number $r$ , the integer powers $r^{k}$ (Integer powers $a^{m}$ ), and the partial sums $s_{n} = \sum_{k < n} r^{k}$ of $\sum r^{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Factorisation of a difference of powers: for $a, b \in R$ and natural $n \geq 1$ , $b^{n} - a^{n} = (b - a) \sum_{k = 0}^{n - 1} a^{k} b^{n - 1 - k}$ (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate).

[L2]

For $∣ r ∣ < 1$ the sequence $(r^{k})$ is null, that is $r^{k} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L3]

Algebra of limits: sums, differences and quotients of convergent sequences converge to the corresponding combination, the quotient rule requiring a nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L4]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , and $∣ x ∣ = 0$ exactly when $x = 0$ ; also $∣ 1 ∣ = 1$ , since $1 > 0$ (Basic properties of the absolute value).

[L5]

Powers and order: $a^{0} = 1$ for every $a$ ; if $a \geq 1$ and $n \geq 1$ then $a^{n} \geq a \geq 1$ ; and $1^{n} = 1$ for every $n$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Integer powers $a^{m}$ ).

[L6]

The principle of induction (The principle of mathematical induction).

[L7]

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

[L8]

Notation of Finite sums and finite products, by recursion: $\sum_{k = 0}^{n - 1} x_{k}$ is $\sum_{k < n} x_{k}$ , and the empty sum $\sum_{k < 0} x_{k}$ is $0$ .

Proof

technique · cases

1.1

Assume $∣ r ∣ < 1$ .

assume-case lt

1.2

Assume instead $∣ r ∣ \geq 1$ .

assume-case ge

1.3

For every natural $n \geq 1$ , applying [L1] with $b = 1$ and $a = r$ gives $1 - r^{n} = (1 - r) \sum_{k = 0}^{n - 1} r^{k} \cdot 1^{n - 1 - k} = (1 - r) s_{n}$ , using $1^{m} = 1$ and the notation of [L8].

L1L5L8

1.4

At $n = 0$ the identity $1 - r^{n} = (1 - r) s_{n}$ also holds, both sides being $0$ because $r^{0} = 1$ and $s_{0}$ is the empty sum.

L5L8

2.1

In the case $∣ r ∣ < 1$ we have $r \neq 1$ , since $∣ 1 ∣ = 1$ and $∣ r ∣ < 1$ ; hence $1 - r \neq 0$ .

step 1.1L4algebra

2.2

In the case $∣ r ∣ \geq 1$ , an induction gives $∣ r^{k} ∣ = ∣ r ∣^{k}$ for every $k \in N$ : at $k = 0$ both sides are $1$ , and if $∣ r^{k} ∣ = ∣ r ∣^{k}$ then $∣ r^{k + 1} ∣ = ∣ r^{k} \cdot r ∣ = ∣ r^{k} ∣ ∣ r ∣ = ∣ r ∣^{k} ∣ r ∣ = ∣ r ∣^{k + 1}$ .

step 1.2L4L5L6

2.3

In the case $∣ r ∣ \geq 1$ we get $∣ r ∣^{k} \geq 1$ for every $k \in N$ : at $k = 0$ this reads $1 \geq 1$ , and for $k \geq 1$ it is the comparison $∣ r ∣^{k} \geq ∣ r ∣ \geq 1$ .

step 1.2L5

3.1

In the case $∣ r ∣ < 1$ , dividing by $1 - r \neq 0$ gives $s_{n} = (1 - r^{n}) / (1 - r)$ for every $n \in N$ .

step 2.1step 1.3step 1.4algebra

3.2

In the case $∣ r ∣ \geq 1$ , combining the two previous steps gives $∣ r^{k} - 0 ∣ = ∣ r^{k} ∣ = ∣ r ∣^{k} \geq 1$ for every $k \in N$ .

step 2.2step 2.3

4.1

In the case $∣ r ∣ < 1$ the sequence $(r^{n})$ is null, so $1 - r^{n} \to 1$ and therefore $s_{n} \to 1 / (1 - r)$ , the denominator being the nonzero constant $1 - r$ ; hence $\sum r^{k}$ converges with sum $1 / (1 - r)$ , which is claim 1.

step 1.1step 3.1step 2.1L2L3

4.2

In the case $∣ r ∣ \geq 1$ the sequence $(r^{k})$ does not converge to $0$ , since the rational tolerance $ε = 1$ admits no index $K$ with $∣ r^{k} - 0 ∣ < 1$ for all $k \geq K$ ; so by the term test $\sum r^{k}$ diverges, which is claim 2.

step 3.2L7

5.1

The two cases $∣ r ∣ < 1$ and $∣ r ∣ \geq 1$ exhaust the possibilities, since the order on $R$ is total, so claims 1 and 2 together cover every real $r$ .

step 4.1step 4.2cases-exhaustive∎

Remarks

The divergence half needs no separate treatment of $r = 1$ and $r = - 1$ . Both are covered by $∣ r ∣ \geq 1$ , and the single reason is the same in every case: the terms have absolute value at least $1$ , so they cannot tend to $0$ . For $r = 1$ the partial sums are $s_{n} = n$ and run to $+ \infty$ ; for $r = - 1$ they oscillate between $0$ and $1$ . The theorem says only that neither converges, which is all that "diverges" means here (Series, partial sums, convergence and the sum, divergence, and the tail series).
Why the identity is proved at $n = 0$ separately. Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate requires $n \geq 1$ , since its right-hand side is a sum over $k < n$ of a term involving $b^{n - 1 - k}$ , and $n - 1$ is not a natural number at $n = 0$ . The identity is still true at $n = 0$ , but by inspection of two empty objects rather than by that lemma, and step 1.4 says so rather than letting the reader assume the citation covers it.

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

$\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$

Statement

Let $(b_{k})$ be a sequence of reals and put $a_{k} : = b_{k} - b_{k + 1}$ . Then the partial sums of $\sum a_{k}$ are

$s_{n} = \sum_{k < n} (b_{k} - b_{k + 1}) = b_{0} - b_{n} (n \in N),$

and consequently $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) if and only if $(b_{k})$ converges (Limits and Cauchy sequences of reals), in which case

$\sum_{k = 0}^{\infty} (b_{k} - b_{k + 1}) = b_{0} - \lim_{k} b_{k} .$

For a family from a general starting index $m$ the same statement holds with $b_{0}$ replaced by $b_{m}$ , being this statement applied to the shifted sequence $j \mapsto b_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

Facts & Assumptions

Given: A sequence $(b_{k})$ of reals, the sequence $a_{k} : = b_{k} - b_{k + 1}$ , and the partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Telescoping of finite sums: $\sum_{k < n} (c_{k + 1} - c_{k}) = c_{n} - c_{0}$ for every sequence $(c_{k})$ of reals and every $n \in N$ (Laws of finite sums and finite products).

[L2]

Algebra of limits: if $x_{n} \to x$ and $y_{n} \to y$ then $x_{n} - y_{n} \to x - y$ (Algebra of limits: sums, scalar multiples, products and quotients).

[L3]

A constant sequence with value $c$ converges to $c$ (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Apply [L1] to the sequence $c_{k} : = - b_{k}$ : it gives $\sum_{k < n} (- b_{k + 1} - (- b_{k})) = - b_{n} - (- b_{0})$ , that is $\sum_{k < n} (b_{k} - b_{k + 1}) = b_{0} - b_{n}$ .

L1algebra

2.1

So $s_{n} = b_{0} - b_{n}$ for every $n \in N$ , and equivalently $b_{n} = b_{0} - s_{n}$ .

step 1.1algebra

3.1

Suppose $(b_{k})$ converges, say $b_{n} \to B$ ; then $s_{n} = b_{0} - b_{n} \to b_{0} - B$ , so $\sum a_{k}$ converges with sum $b_{0} - B$ .

step 2.1L2L3

3.2

Suppose conversely that $\sum a_{k}$ converges, say $s_{n} \to S$ ; then $b_{n} = b_{0} - s_{n} \to b_{0} - S$ , so $(b_{k})$ converges.

step 2.1L2L3

4.1

The two implications give the stated equivalence, and in the convergent case step 3.1 gives the sum $b_{0} - \lim_{k} b_{k}$ .

step 3.1step 3.2∎

Remarks

The first term of $(b_{k})$ survives and the rest cancel. That is the whole content, and it is where an off-by-one error is easiest to make: the sum is $b_{0} - \lim b_{k}$ and not $b_{1} - \lim b_{k}$ , because the partial sum $s_{n} = \sum_{k < n}$ starts at $k = 0$ (Series, partial sums, convergence and the sum, divergence, and the tail series). A series written from $k = 1$ telescopes to $b_{1} - \lim b_{k}$ instead.
No sign or monotonicity hypothesis is used. The lemma is an identity between finite sums followed by one limit rule, so it applies to sequences $(b_{k})$ of any sign and is not restricted to the decreasing case in which telescoping series usually appear.

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

A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} \geq 0$ for every $k \in N$ , let $s_{n} = \sum_{k < n} a_{k}$ be its partial sums and let $S = {s_{n} : n \in N}$ be the range of $(s_{n})$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

$(s_{n})$ is nondecreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and $s_{n} \geq 0$ for every $n$ ;
$\sum a_{k}$ converges if and only if $S$ is bounded above (Lower bound, bounded below, bounded set), and in that case $\sum_{k = 0}^{\infty} a_{k} = \sup S,$ so in particular $s_{n} \leq \sum_{k = 0}^{\infty} a_{k}$ for every $n$ ;
if $S$ is not bounded above then $s_{n} \to + \infty$ (Divergence to $+ \infty$ and to $- \infty$ ) and $\sum a_{k}$ diverges.

This is the theorem that makes the nonnegative theory work: for terms of one sign, convergence is a boundedness question and no candidate limit is ever needed. Every comparison test on this page is an application of it.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} \geq 0$ for every $k$ , its partial sums $s_{n} = \sum_{k < n} a_{k}$ , and the range $S = {s_{n} : n \in N}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The recursion clause of the finite sum: $s_{n + 1} = s_{n} + a_{n}$ (Finite sums and finite products, by recursion).

[L2]

Consecutive comparisons suffice for monotonicity: $(x_{k})$ is nondecreasing if and only if $x_{k} \leq x_{k + 1}$ for every $k$ ; and a nondecreasing sequence is bounded below by its first term (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Monotonicity of finite sums: if $a_{k} \geq 0$ for all $k < n$ then $\sum_{k < n} a_{k} \geq 0$ (Laws of finite sums and finite products).

[L4]

A monotone sequence converges if and only if it is bounded, that is if and only if there is $M$ with $∣ x_{k} ∣ \leq M$ for every $k$ (A monotone sequence converges if and only if it is bounded, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

A nondecreasing sequence bounded above converges to the supremum of its range, which exists by the least-upper-bound property (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set).

[L6]

A nondecreasing sequence whose range is not bounded above diverges to $+ \infty$ (A nondecreasing sequence that is not bounded above diverges to $+ \infty$ , Divergence to $+ \infty$ and to $- \infty$ ).

Proof

technique · direct

1.1

For every $n$ , $s_{n + 1} - s_{n} = a_{n} \geq 0$ , so $s_{n} \leq s_{n + 1}$ and $(s_{n})$ is nondecreasing.

givenL1L2

1.2

For every $n$ , $s_{n} = \sum_{k < n} a_{k} \geq 0$ , all terms being nonnegative.

givenL3

2.1

Claim 1 is steps 1.1 and 1.2 together.

step 1.1step 1.2

2.2

Since $s_{n} \geq 0$ we have $∣ s_{n} ∣ = s_{n}$ , so $(s_{n})$ is bounded in the sense of [L4] if and only if $S$ is bounded above.

step 1.2L4

3.1

By [L4] applied to the monotone sequence $(s_{n})$ , the series converges if and only if $(s_{n})$ is bounded, hence if and only if $S$ is bounded above.

step 1.1step 2.2L4

4.1

If $S$ is bounded above then $(s_{n})$ converges to $\sup S$ , so $\sum a_{k}$ converges with sum $\sup S$ ; and since $\sup S$ is an upper bound of $S$ , $s_{n} \leq \sup S$ for every $n$ .

step 1.1step 3.1L5

4.2

If $S$ is not bounded above then $s_{n} \to + \infty$ , and by step 3.1 the series diverges.

step 1.1step 3.1L6

5.1

The equivalence and the identification of the sum as the supremum together make claim 2, and the divergence statement is claim 3.

step 3.1step 4.1step 4.2∎

Remarks

"Bounded" and "bounded above" coincide here, and only here. The equivalence used in step 2.2 rests on $s_{n} \geq 0$ , which rests on every term being nonnegative. For a series with terms of both signs the partial sums can be bounded above and still fail to converge, so nothing in this theorem survives the loss of the sign hypothesis. That failure is exhibited by Two series with $a_{k} \leq b_{k}$ for all $k$ , $\sum b_{k}$ convergent and $\sum a_{k}$ divergent, when the terms may be negative ↗ on the companion page.
Claim 3 is a strictly stronger statement than "diverges". Divergence alone permits oscillation (Series, partial sums, convergence and the sum, divergence, and the tail series); for nonnegative terms it cannot occur, and the partial sums necessarily run to $+ \infty$ . This is what licenses the phrase "the series diverges to $+ \infty$ " for nonnegative terms, and it is what the Abel-Dini theorem later on this page uses.
This criterion is the monotone convergence property, worn differently. The proof above is monotone convergence for $R$ applied to the nondecreasing sequence of partial sums, and nothing is lost going back the other way. Given a nondecreasing sequence $(x_{n})$ of reals, put $y_{n} : = x_{n} - x_{0} \geq 0$ and let $a_{0} : = 0$ , $a_{k + 1} : = y_{k + 1} - y_{k} \geq 0$ ; then $s_{n + 1} = y_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), the partial sums are bounded exactly when $(x_{n})$ is bounded above, and claim 1 returns the convergence of $(y_{n})$ and so of $(x_{n})$ . Testing boundedness of partial sums is therefore not a device special to series. Read in the vocabulary of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness it is the property (MCT), which in an arbitrary ordered field already forces the Archimedean property on its own (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis) and with it the least-upper-bound property (The monotone convergence property plus the Archimedean property imply the least-upper-bound property). The translation just given is carried out in $R$ , since Series, partial sums, convergence and the sum, divergence, and the tail series is stated for sequences of reals and this library defines no series over a general ordered field.

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

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 $\sum b_{k}$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals and suppose there is $K \in N$ with

$0 \leq a_{k} \leq b_{k} for all k \geq K .$

Then:

if $\sum b_{k}$ converges then $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series);
if $\sum a_{k}$ diverges then $\sum b_{k}$ diverges.

The same statement holds verbatim for series with a general starting index $m$ , applied to the shifted sequences of Series, partial sums, convergence and the sum, divergence, and the tail series.

The hypothesis is on the terms from some index on, not on all of them: finitely many terms of either sequence may violate it, or be negative, without affecting the conclusion. What may not be dropped is nonnegativity of $(a_{k})$ from that index on.

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals and $K \in N$ with $0 \leq a_{k} \leq b_{k}$ for all $k \geq K$ ; the partial sums $α_{j} = \sum_{i < j} a_{K + i}$ and $β_{j} = \sum_{i < j} b_{K + i}$ of the $K$ -th tail series (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Monotonicity of finite sums: if $x_{i} \leq y_{i}$ for all $i < j$ then $\sum_{i < j} x_{i} \leq \sum_{i < j} y_{i}$ (Laws of finite sums and finite products).

[L2]

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

[L3]

For a series of nonnegative terms: it converges if and only if the range of its partial sums is bounded above, and in the convergent case every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

For every $i \in N$ the index $K + i$ is at least $K$ , so $0 \leq a_{K + i} \leq b_{K + i}$ ; in particular both tail series have nonnegative terms.

given

1.2

Assume $\sum b_{k}$ converges. Then its $K$ -th tail series $\sum_{k \geq K} b_{k}$ converges.

givenL2

2.1

By monotonicity of finite sums, $α_{j} \leq β_{j}$ for every $j \in N$ .

step 1.1L1

2.2

That tail series has nonnegative terms, so its partial sums satisfy $β_{j} \leq B$ for every $j$ , where $B$ is its sum.

step 1.2step 1.1L3

3.1

Hence $α_{j} \leq β_{j} \leq B$ for every $j$ , so the range of $(α_{j})$ is bounded above by $B$ .

step 2.1step 2.2

4.1

The tail series $\sum_{k \geq K} a_{k}$ has nonnegative terms and partial sums bounded above, so it converges.

step 3.1step 1.1L3

5.1

Therefore $\sum a_{k}$ converges, which is claim 1.

step 4.1L2

6.1

Claim 2 is the contrapositive of claim 1: if $\sum a_{k}$ diverges then $\sum b_{k}$ cannot converge.

step 5.1∎

Remarks

Both nonnegativity hypotheses are used, and in different places. $a_{k} \geq 0$ is what lets convergence of $\sum_{k \geq K} a_{k}$ be read off from boundedness of its partial sums, and $b_{k} \geq a_{k} \geq 0$ is what makes the sum of $\sum_{k \geq K} b_{k}$ an upper bound for the partial sums $β_{j}$ . Drop the sign hypothesis and the theorem is false, not merely unproved; the companion page exhibits a pair with $a_{k} \leq b_{k}$ for every $k$ , $\sum b_{k}$ convergent and $\sum a_{k}$ divergent.
The comparison is with a series, not with a limit. No quotient $a_{k} / b_{k}$ appears and no $b_{k}$ is required to be nonzero, which is what distinguishes this test from the limit comparison test proved next.

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

For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals with $a_{k} > 0$ and $b_{k} > 0$ for every $k \in N$ , and put $q_{k} : = a_{k} / b_{k}$ . Then:

if $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for a real $L > 0$ (Limits and Cauchy sequences of reals), then $\sum a_{k}$ converges if and only if $\sum b_{k}$ converges;
if $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ , then convergence of $\sum b_{k}$ implies convergence of $\sum a_{k}$ ; equivalently, divergence of $\sum a_{k}$ implies divergence of $\sum b_{k}$ ;
if $(q_{k})$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ), then convergence of $\sum a_{k}$ implies convergence of $\sum b_{k}$ ; equivalently, divergence of $\sum b_{k}$ implies divergence of $\sum a_{k}$ .

In each clause the convergence of $(q_{k})$ , or its divergence to $+ \infty$ , is part of the hypothesis, so the symbol $\lim_{k} q_{k}$ denotes wherever it is written (A sequence has at most one limit).

Neither implication in claim 2 can be reversed, and by symmetry neither can the one in claim 3; the companion page exhibits a pair with $\lim_{k} q_{k} = 0$ , $\sum a_{k}$ convergent and $\sum b_{k}$ divergent.

For families from a general starting index $m$ the statement is the same, applied to the shifted sequences $j \mapsto a_{j + m}$ and $j \mapsto b_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

On the third regime. " $L = \infty$ " is written here as divergence of $(q_{k})$ to $+ \infty$ in the sense of Divergence to $+ \infty$ and to $- \infty$ , and never as a limit equation with an infinite right-hand side. A sequence diverging to $+ \infty$ has no limit in $R$ , and this library does not write $\lim q_{k} = + \infty$ .

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals with $a_{k} > 0$ and $b_{k} > 0$ for every $k$ , the quotients $q_{k} = a_{k} / b_{k}$ , and the assumption that one of the three regimes of the Statement holds: $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for some real $L > 0$ ; or $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ ; or $(q_{k})$ diverges to $+ \infty$ (A sequence has at most one limit).

[L1]

Convergence to $x$ means: for every rational $ε > 0$ there is $K$ with $∣ x_{k} - x ∣ < ε$ for all $k \geq K$ ; and the same holds for every real $ε > 0$ , since every real $ε > 0$ exceeds some rational $1 / n$ with $n \geq 1$ natural (Limits and Cauchy sequences of reals, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

$q_{k} \to + \infty$ means: for every real $M$ there is $K$ with $q_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[L3]

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

[L4]

For $c \neq 0$ : $\sum c x_{k}$ converges if and only if $\sum x_{k}$ converges (Convergent series add and scale termwise).

[L5]

Since $q_{k} = a_{k} / b_{k}$ and $b_{k} > 0$ , the field laws give $a_{k} = q_{k} b_{k}$ . Multiplication by a positive scalar preserves strict inequalities; the non-strict form follows by adjoining the equality case (Field, Sign rules for products and monotonicity of multiplication).

Proof

technique · cases

1.1

Assume $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for a real $L > 0$ .

assume-case pos

1.2

Assume instead $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ .

assume-case zero

1.3

Assume instead $(q_{k})$ diverges to $+ \infty$ .

assume-case inf

2.1

In the case $\lim_{k} q_{k} = L > 0$ , apply [L1] with the real tolerance $L / 2 > 0$ : there is $K$ with $∣ q_{k} - L ∣ < L / 2$ , hence $L / 2 < q_{k} < 3 L / 2$ , for all $k \geq K$ .

step 1.1L1choose

2.2

In the case $\lim_{k} q_{k} = 0$ , apply [L1] with the rational tolerance $1$ : there is $K$ with $∣ q_{k} ∣ < 1$ , hence $q_{k} < 1$ , for all $k \geq K$ .

step 1.2L1choose

2.3

In the case $q_{k} \to + \infty$ , apply [L2] with $M = 1$ : there is $K$ with $q_{k} > 1$ for all $k \geq K$ .

step 1.3L2choose

3.1

In the case $\lim_{k} q_{k} = L > 0$ , multiplying by $b_{k} > 0$ turns step 2.1 into $(L / 2) b_{k} < a_{k} < (3 L / 2) b_{k}$ for all $k \geq K$ , and all three quantities are positive.

step 2.1L5algebra

3.2

In the case $\lim_{k} q_{k} = 0$ , multiplying by $b_{k} > 0$ turns step 2.2 into $0 < a_{k} < b_{k}$ for all $k \geq K$ .

step 2.2L5algebra

3.3

In the case $q_{k} \to + \infty$ , multiplying by $b_{k} > 0$ turns step 2.3 into $0 < b_{k} < a_{k}$ for all $k \geq K$ .

step 2.3L5algebra

4.1

In the case $L > 0$ : if $\sum b_{k}$ converges then so does $\sum (3 L / 2) b_{k}$ , and $0 \leq a_{k} \leq (3 L / 2) b_{k}$ for $k \geq K$ , so $\sum a_{k}$ converges.

step 3.1L3L4

4.2

In the case $L > 0$ : if $\sum a_{k}$ converges then, since $0 \leq (L / 2) b_{k} \leq a_{k}$ for $k \geq K$ , the series $\sum (L / 2) b_{k}$ converges, and $L / 2 \neq 0$ gives convergence of $\sum b_{k}$ .

step 3.1L3L4

4.3

In the case $\lim_{k} q_{k} = 0$ : $0 \leq a_{k} \leq b_{k}$ for $k \geq K$ , so convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and the contrapositive is the divergence form.

step 3.2L3

4.4

In the case $q_{k} \to + \infty$ : $0 \leq b_{k} \leq a_{k}$ for $k \geq K$ , so convergence of $\sum a_{k}$ gives convergence of $\sum b_{k}$ , and the contrapositive is the divergence form.

step 3.3L3

5.1

The two implications in the case $L > 0$ are the two directions of claim 1, and the remaining two cases give claims 2 and 3. The three assumed regimes are the cases of the disjunction in the Given, and they exhaust it, so every instance of the theorem is covered: outside those three regimes each of the three implications is vacuous, its hypothesis being false.

step 4.1step 4.2step 4.3step 4.4cases-exhaustive∎

Remarks

Why the three regimes are treated as one proof. The Statement is a conjunction of three implications, each with its own hypothesis on $(q_{k})$ . Fixing the two sequences and arguing by cases on which regime holds proves all three at once, and costs nothing: if none of the regimes holds, every one of the three implications is vacuously true.
Positivity of $(b_{k})$ is needed twice. It is what makes $q_{k}$ defined at all, and it is what lets an inequality between the $q_{k}$ be multiplied through to an inequality between the $a_{k}$ and the $b_{k}$ without reversing. Positivity of $(a_{k})$ is what supplies the lower bound $0$ that the direct comparison test requires.
The limit is only used through an eventual two-sided estimate. No step needs the exact value of $L$ , only that $q_{k}$ is eventually trapped strictly between two positive multiples of it. That is why the test still works when the quotients merely stay between two positive constants, and why the hypothesis $\lim_{k} q_{k} = L$ is stronger than what the proof consumes.

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

For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges

Statement

Let $(a_{k})_{k \geq 1}$ be a family from $1$ (Series, partial sums, convergence and the sum, divergence, and the tail series) with

$a_{k} \geq 0 (k \geq 1) and a_{j} \geq a_{k} whenever 1 \leq j \leq k .$

Then

$\sum_{k \geq 1} a_{k} converges ⟺ \sum_{j \geq 0} 2^{j} a_{2^{j}} converges .$

Every term of the condensed series is defined, because $2^{j} \geq 1$ for every $j \in N$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ), and the condensed series starts at $j = 0$ , its first term being $2^{0} a_{2^{0}} = a_{1}$ .

The monotonicity hypothesis is equivalent to the consecutive form $a_{k + 1} \leq a_{k}$ for every $k \geq 1$ , since it says that the sequence $b_{i} : = a_{i + 1}$ , $i \in N$ , is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences). It cannot be dropped: the companion page exhibits a nonnegative non-monotone family for which the two series behave differently.

Facts & Assumptions

Given: A family $(a_{k})_{k \geq 1}$ of reals with $a_{k} \geq 0$ for $k \geq 1$ and $a_{j} \geq a_{k}$ whenever $1 \leq j \leq k$ ; the partial sums $A_{n} = \sum_{k = 1}^{n} a_{k}$ of $\sum_{k \geq 1} a_{k}$ , with $A_{0} = 0$ ; and the partial sums $C_{n} = \sum_{j < n} 2^{j} a_{2^{j}}$ of the condensed series (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Splitting of finite sums, and the meaning of a sum with general bounds: for $m \leq n$ , $\sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{m} a_{k} + \sum_{k = m + 1}^{n} a_{k}$ , and $\sum_{k = p}^{q} a_{k} = \sum_{t < q + 1 - p} a_{p + t}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L2]

Monotonicity and scaling of finite sums: if $x_{t} \leq y_{t}$ for all $t < d$ then $\sum_{t < d} x_{t} \leq \sum_{t < d} y_{t}$ ; and a constant sum is $\sum_{t < d} λ = d λ$ (Laws of finite sums and finite products).

[L3]

Powers of $2$ : $2^{j} \geq 1$ for every $j \in N$ , $2^{j + 1} = 2 \cdot 2^{j}$ , and $2^{j} < 2^{j + 1}$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

The principle of induction (The principle of mathematical induction).

[L5]

For a series of nonnegative terms: its partial sums are nondecreasing, it converges if and only if the range of its partial sums is bounded above, and in the convergent case every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

Every term of the condensed series is nonnegative, since $2^{j} \geq 1 > 0$ and $a_{2^{j}} \geq 0$ ; and every $A_{n}$ , every $C_{n}$ is nonnegative, both series having nonnegative terms.

givenL3L5

1.2

For every $n \in N$ the block from $2^{n}$ to $2^{n + 1} - 1$ is $\sum_{k = 2^{n}}^{2^{n + 1} - 1} a_{k} = \sum_{t < 2^{n}} a_{2^{n} + t}$ , since the number of terms is $(2^{n + 1} - 1) + 1 - 2^{n} = 2^{n}$ .

L1L3algebra

1.3

For every $n \in N$ the block from $2^{n} + 1$ to $2^{n + 1}$ is $\sum_{k = 2^{n} + 1}^{2^{n + 1}} a_{k} = \sum_{t < 2^{n}} a_{2^{n} + 1 + t}$ , since the number of terms is $2^{n + 1} + 1 - (2^{n} + 1) = 2^{n}$ .

L1L3algebra

1.4

Index growth: $n \leq 2^{n} - 1$ for every $n \in N$ , by induction on $n$ . At $n = 0$ this reads $0 \leq 0$ ; and if $n \leq 2^{n} - 1$ , that is $n + 1 \leq 2^{n}$ , then $n + 1 \leq 2^{n} \leq 2 \cdot 2^{n} - 1 = 2^{n + 1} - 1$ , the second inequality being $2^{n} \geq 1$ .

L3L4algebra

2.1

In the first block every index satisfies $2^{n} \leq 2^{n} + t$ , so $a_{2^{n} + t} \leq a_{2^{n}}$ , and therefore $\sum_{k = 2^{n}}^{2^{n + 1} - 1} a_{k} \leq \sum_{t < 2^{n}} a_{2^{n}} = 2^{n} a_{2^{n}}$ .

step 1.2givenL2

2.2

In the second block every index satisfies $2^{n} + 1 + t \leq 2^{n + 1}$ for $t < 2^{n}$ , so $a_{2^{n} + 1 + t} \geq a_{2^{n + 1}}$ , and therefore $\sum_{k = 2^{n} + 1}^{2^{n + 1}} a_{k} \geq \sum_{t < 2^{n}} a_{2^{n + 1}} = 2^{n} a_{2^{n + 1}}$ .

step 1.3givenL2L3

2.3

Suppose the condensed series converges, with sum $C$ ; then $C_{n} \leq C$ for every $n$ .

givenstep 1.1L5

2.4

Suppose conversely that $\sum_{k \geq 1} a_{k}$ converges, with sum $A$ ; then $A_{n} \leq A$ for every $n$ .

givenstep 1.1L5

3.1

Upper estimate: $A_{2^{n} - 1} \leq C_{n}$ for every $n \in N$ , by induction on $n$ . At $n = 0$ both sides are $0$ , since $2^{0} - 1 = 0$ and $C_{0}$ is the empty sum; and if $A_{2^{n} - 1} \leq C_{n}$ , then splitting at $2^{n} - 1$ gives $A_{2^{n + 1} - 1} = A_{2^{n} - 1} + \sum_{k = 2^{n}}^{2^{n + 1} - 1} a_{k} \leq C_{n} + 2^{n} a_{2^{n}} = C_{n + 1}$ .

step 2.1L1L3L4

3.2

Lower estimate: $2 A_{2^{n}} \geq \sum_{j = 1}^{n} 2^{j} a_{2^{j}}$ for every $n \in N$ , by induction on $n$ . At $n = 0$ the right-hand side is the empty sum $0$ and the left-hand side is $2 a_{1} \geq 0$ ; and if the inequality holds at $n$ , then splitting at $2^{n}$ gives $A_{2^{n + 1}} = A_{2^{n}} + \sum_{k = 2^{n} + 1}^{2^{n + 1}} a_{k} \geq A_{2^{n}} + 2^{n} a_{2^{n + 1}}$ , whence $2 A_{2^{n + 1}} \geq 2 A_{2^{n}} + 2^{n + 1} a_{2^{n + 1}} \geq \sum_{j = 1}^{n + 1} 2^{j} a_{2^{j}}$ .

step 2.2step 1.1L1L3L4

4.1

For every $N \in N$ we have $N \leq 2^{N} - 1$ , so $A_{N} \leq A_{2^{N} - 1} \leq C_{N} \leq C$ , the first inequality because the partial sums $A_{n}$ are nondecreasing.

step 1.4step 3.1step 2.3step 1.1L5

4.2

For every $n \in N$ , $\sum_{j = 1}^{n} 2^{j} a_{2^{j}} \leq 2 A_{2^{n}} \leq 2 A$ , and splitting the condensed partial sum at $j = 1$ gives $C_{n + 1} = a_{1} + \sum_{j = 1}^{n} 2^{j} a_{2^{j}} \leq a_{1} + 2 A$ .

step 3.2step 2.4L1L3

5.1

So the partial sums of $\sum_{k \geq 1} a_{k}$ are bounded above by $C$ , and that series converges.

step 4.1L5

5.2

Also $C_{0} = 0 \leq a_{1} + 2 A$ , so every condensed partial sum is at most $a_{1} + 2 A$ , and the condensed series converges.

step 4.2step 1.1L5

6.1

The two implications just established combine, so the two series converge or diverge together.

step 5.1step 5.2∎

Remarks

What monotonicity buys, in one sentence. It lets a block of $2^{n}$ consecutive terms be squeezed between $2^{n}$ copies of its last term and $2^{n}$ copies of its first, which is exactly the pair of estimates in steps 2.1 and 2.2. Without it a block carries no information about any single term in it, and the two series decouple entirely.
The factor $2$ in the lower estimate is not an artefact. The blocks used for the two estimates are different: the upper estimate groups $2^{n}, \dots, 2^{n + 1} - 1$ and the lower estimate groups $2^{n} + 1, \dots, 2^{n + 1}$ , and the second grouping produces $2^{n} a_{2^{n + 1}}$ , which is half of the condensed term $2^{n + 1} a_{2^{n + 1}}$ . Since only boundedness of the partial sums is at stake, a constant factor is harmless.
Base $2$ is a choice, not a necessity. The same argument with blocks of length $c^{n}$ gives the analogous test for any integer $c \geq 2$ . Base $2$ is taken here because it is the one every later application uses, and because the arithmetic of $2^{n + 1} = 2 \cdot 2^{n}$ keeps the induction free of extra bookkeeping.

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

For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$

Statement

Let $p \in Q$ with $p > 0$ . For a natural number $k \geq 1$ write $ι (k) = k \cdot 1_{R}$ for the canonical natural, which is positive (Canonical naturals are positive and strictly increasing), and write $k^{p} : = ι (k)^{p}$ for its rational power (Rational powers $a^{r}$ of a positive base). Then

$\sum_{k \geq 1} \frac{1}{k^{p}} converges ⟺ p > 1.$

In particular the harmonic series $\sum_{k \geq 1} 1 / k$ diverges, at $p = 1$ , and $\sum_{k \geq 1} 1 / k^{2}$ converges, at $p = 2$ .

The index range is not cosmetic. The series starts at $k = 1$ because $1 / 0^{p}$ is undefined: Rational powers $a^{r}$ of a positive base gives $0^{p} = 0$ for rational $p > 0$ , and $0$ has no inverse. Sequences here are functions on $N$ and $N$ contains $0$ (Series, partial sums, convergence and the sum, divergence, and the tail series), so the object named above is a series from the starting index $1$ in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series, not a series of a sequence on $N$ .

The exponent is rational, and that is a limitation of this page. Rational powers of a positive base are what Rational powers $a^{r}$ of a positive base supplies; real exponents require the exponential and the logarithm, which this library develops later. The statement above is therefore the full $p$ -series theorem for every exponent this page can name.

Facts & Assumptions

Given: A rational $p > 0$ and the family $a_{k} : = 1 / k^{p} = ι (k)^{- p}$ , defined for naturals $k \geq 1$ (Rational powers $a^{r}$ of a positive base, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

Rational powers of a positive base are positive, and $a^{r + s} = a^{r} a^{s}$ , $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ for $a > 0$ and rationals $r, s$ (Laws of rational exponents).

[L2]

Monotonicity of rational powers: for rational $t > 0$ and $0 < a < b$ one has $a^{t} < b^{t}$ ; and for $a > 1$ and rationals $r < s$ one has $a^{r} < a^{s}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L3]

The integer power and the rational power agree at an integer exponent: for $a > 0$ and $n \in Z$ , $a^{n}$ read as in Integer powers $a^{m}$ equals $a^{n}$ read as in Rational powers $a^{r}$ of a positive base, since $n = n / 1$ and $a^{1 / 1} = a$ (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). In particular $a^{0} = 1$ .

[L4]

Reciprocation reverses the order on the positives: $0 < a < b$ implies $0 < 1 / b < 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Condensation: for a family $(x_{k})_{k \geq 1}$ that is nonnegative and nonincreasing, $\sum_{k \geq 1} x_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} x_{2^{j}}$ converges (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L6]

The geometric series: $\sum_{j \geq 0} r^{j}$ converges if and only if $∣ r ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Basic properties of the absolute value).

[L7]

The canonical naturals are positive and order preserving: $0 < ι (1) \leq ι (j) < ι (k)$ for naturals $1 \leq j < k$ , and $ι (2) = 2 > 1$ (Canonical naturals are positive and strictly increasing).

Proof

technique · direct

1.1

For every natural $k \geq 1$ the base $ι (k)$ is positive, so $a_{k} = ι (k)^{- p}$ is defined and positive; in particular the family is nonnegative.

givenL7L1

1.2

For naturals $1 \leq j < k$ we have $0 < ι (j) < ι (k)$ , hence $ι (j)^{p} < ι (k)^{p}$ since $p > 0$ , hence $a_{j} = 1 / ι (j)^{p} > 1 / ι (k)^{p} = a_{k}$ ; and for $j = k$ the two are equal. So $a_{j} \geq a_{k}$ whenever $1 \leq j \leq k$ .

givenL7L2L4L1

1.3

For every $j \in N$ the base $2^{j}$ is positive and, reading the exponent $j$ as a rational, $2^{j} a_{2^{j}} = 2^{j} (2^{j})^{- p} = 2^{j} \cdot 2^{- j p} = 2^{j - j p} = 2^{(1 - p) j} = (2^{1 - p})^{j}$ .

L1L3L7algebra

1.4

Since $2 > 1$ , the map $t \mapsto 2^{t}$ is strictly increasing on $Q$ and $2^{0} = 1$ ; hence $r = 2^{1 - p} < 1 = 2^{0}$ holds exactly when $1 - p < 0$ , that is exactly when $p > 1$ .

L2L3L7

2.1

Condensation applies to $(a_{k})_{k \geq 1}$ : $\sum_{k \geq 1} a_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} a_{2^{j}}$ converges.

step 1.1step 1.2L5

2.2

So the condensed series is the geometric series $\sum_{j \geq 0} r^{j}$ with $r : = 2^{1 - p}$ , and $r > 0$ , so $∣ r ∣ = r$ .

step 1.3L1L3

3.1

By the geometric series theorem, $\sum_{j \geq 0} r^{j}$ converges if and only if $r < 1$ .

step 2.2L6

4.1

Chaining the three equivalences: $\sum_{k \geq 1} 1 / k^{p}$ converges $⟺$ the condensed series converges $⟺$ $r < 1$ $⟺$ $p > 1$ .

step 2.1step 2.2step 3.1step 1.4∎

Remarks

Where the threshold comes from. Condensation turns the $p$ -series into a geometric series of ratio $2^{1 - p}$ , and the geometric threshold $r = 1$ pulls back to $p = 1$ . Nothing about the number $1$ is special to the $p$ -series; it is the exponent at which the condensed terms stop shrinking.
At $p = 1$ the condensed series is $\sum_{j \geq 0} 1$ . Its terms do not tend to $0$ , so it diverges, and with it the harmonic series. That instance is worked out on the companion page, together with the older block argument that does not use condensation at all.
Only rational exponents are covered, and the gap is real. For irrational $p$ the expression $k^{p}$ has no meaning in this library yet, so the statement is not merely unproved there, it is unstatable. The same limitation is what keeps the Bertrand-type series $\sum 1 / (k (\log k)^{p})$ off this page entirely, the logarithm not being available.

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

Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing

Statement

Let $(a_{k})_{k \geq 1}$ be a family of reals from the starting index $1$ (Series, partial sums, convergence and the sum, divergence, and the tail series), put

$ρ_{k} : = ∣ a_{k + 1} ∣^{1 / (k + 1)} (k \in N), ρ : = {lim sup}_{k} ρ_{k} \in \overline{R},$

and note that $ρ$ exists for every such family, with no hypothesis whatever (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ). Then:

if $ρ < 1$ then $\sum_{k \geq 1} ∣ a_{k} ∣$ converges, and hence $\sum_{k \geq 1} a_{k}$ converges as well;
if $ρ > 1$ then $\sum_{k \geq 1} a_{k}$ diverges;
if $ρ = 1$ neither conclusion follows: $\sum_{k \geq 1} 1 / k$ diverges, $\sum_{k \geq 1} 1 / k^{2}$ converges, and both have $ρ = 1$ .

The root family is shifted, and that is forced. The classical expression $∣ a_{n} ∣^{1 / n}$ is meaningful only for $n \geq 1$ , since $1 / 0$ is not a rational number (Rational powers $a^{r}$ of a positive base), while sequences here are functions on $N$ and $N$ contains $0$ . So the roots are written $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , which is $∣ a_{n} ∣^{1 / n}$ reindexed by $n = k + 1$ , exactly the convention of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ . Every $ρ_{k}$ is defined, including where $a_{k + 1} = 0$ , by the supplementary clause of Rational powers $a^{r}$ of a positive base.

What claim 1 does and does not say. The comparison with a geometric series delivers convergence of the series of absolute values; that $\sum_{k \geq 1} a_{k}$ itself converges is a separate step, and it is supplied by If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges earlier on this page. Nothing here identifies the sum, and nothing here says anything about rearranging the series, which is taken up later in this track.

Facts & Assumptions

Given: A family $(a_{k})_{k \geq 1}$ of reals, the roots $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ for $k \in N$ , the tail suprema $s_{n} = \sup {ρ_{k} : k \geq n}$ taken in $\overline{R}$ , and $ρ = {lim sup}_{k} ρ_{k} = \inf {s_{n} : n \in N}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, and the extended order is total (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). In particular $ρ \leq s_{n}$ for every $n$ , and $ρ_{k} \leq s_{n}$ for every $k \geq n$ ; a real $t$ with $ρ < t$ fails to be a lower bound of ${s_{n}}$ , and a real $u$ with $s_{n} > u$ fails to be an upper bound of ${ρ_{k} : k \geq n}$ .

[L2]

Both quantities exist for every sequence, bounded or not (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L3]

Roots and powers: for $x \geq 0$ and natural $n \geq 1$ , $x^{1 / n} \geq 0$ and $(x^{1 / n})^{n} = x$ ; on the nonnegatives $y \mapsto y^{n}$ is strictly increasing for $n \geq 1$ ; and $1^{n} = 1$ (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, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

Absolute value: $∣ x ∣ \geq 0$ for every real $x$ (Basic properties of the absolute value).

[L5]

The geometric series $\sum_{j \geq 0} t^{j}$ converges when $∣ t ∣ < 1$ , and a series converges if and only if each of its tail series converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L6]

Direct comparison, in the form for families from a general starting index: if $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum y_{k}$ converges then $\sum x_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

If a series converges then its terms tend to $0$ ; contrapositively, terms not tending to $0$ force divergence (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L8]

$1 \leq n^{1 / n}$ for every natural $n \geq 1$ , and the sequence $(k + 1)^{1 / (k + 1)}$ converges to $1$ ( $n^{1 / n} \to 1$ ); a sequence converging to a real $c$ has $lim sup = lim inf = c$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ ); products and quotients of convergent sequences converge, the quotient requiring nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients).

[L9]

Laws of rational exponents on a positive base: $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ and $a^{r} > 0$ ; and for rational $t > 0$ , $a > 1$ implies $a^{t} > 1$ (Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L10]

For rational $p > 0$ , $\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[L11]

If $\sum ∣ x_{j} ∣$ converges then $\sum x_{j}$ converges; for a family from the starting index $1$ this is the same statement applied to the shifted sequence $j \mapsto a_{j + 1}$ , whose series is $\sum_{k \geq 1} a_{k}$ and whose absolute-value series is $\sum_{k \geq 1} ∣ a_{k} ∣$ (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · cases

1.1

Assume $ρ < 1$ .

assume-case lt

1.2

Assume instead $ρ > 1$ .

assume-case gt

1.3

Assume instead $ρ = 1$ .

assume-case one

2.1

Every $ρ_{k}$ is a nonnegative real, so each $s_{n} \geq ρ_{n} \geq 0$ and hence $0$ is a lower bound of ${s_{n}}$ , giving $ρ \geq 0$ ; combined with the case hypothesis $ρ < 1$ this puts $ρ$ strictly between the reals $0$ and $1$ , so $ρ$ is a real number.

step 1.1L1L2L3L4

2.2

In the case $ρ > 1$ , the value $ρ$ is a lower bound of ${s_{n}}$ , so $s_{n} \geq ρ > 1$ for every $n \in N$ .

step 1.2L1

2.3

In the case $ρ = 1$ , take first $b_{k} : = 1 / k$ for $k \geq 1$ . Its root family is $(1 / (k + 1))^{1 / (k + 1)} = 1 / (k + 1)^{1 / (k + 1)}$ , and since $(k + 1)^{1 / (k + 1)} \to 1$ with every term at least $1$ , the quotient rule gives convergence to $1$ , so the limit superior of the root family is $1$ ; and $\sum_{k \geq 1} 1 / k$ diverges, being the case $p = 1$ .

step 1.3L8L9L10

2.4

In the case $ρ = 1$ , take next $c_{k} : = 1 / k^{2}$ for $k \geq 1$ . Its root family is $((k + 1)^{- 2})^{1 / (k + 1)} = ((k + 1)^{1 / (k + 1)})^{- 2}$ , which converges to $1^{- 2} = 1$ by the product and quotient rules, so again the limit superior of the root family is $1$ ; and $\sum_{k \geq 1} 1 / k^{2}$ converges, being the case $p = 2$ .

step 1.3L8L9L10

3.1

In the case $ρ < 1$ put $t : = (ρ + 1) / 2$ , a real number with $0 \leq ρ < t < 1$ ; since $t$ is not a lower bound of ${s_{n}}$ there is $N \in N$ with $s_{N} < t$ .

step 2.1L1choose

3.2

In the case $ρ > 1$ , for each $n$ the real $1$ is not an upper bound of ${ρ_{k} : k \geq n}$ , so there is $k \geq n$ with $ρ_{k} > 1$ .

step 2.2L1

3.3

So at $ρ = 1$ one family gives a divergent series and another a convergent one, and neither of the two conclusions can be drawn, which is claim 3.

step 2.3step 2.4

4.1

In the case $ρ < 1$ , for every $k \geq N$ we have $ρ_{k} \leq s_{N} < t$ , and raising both nonnegative sides to the power $k + 1 \geq 1$ gives $∣ a_{k + 1} ∣ = (ρ_{k})^{k + 1} < t^{k + 1}$ .

step 3.1L1L3

4.2

In the case $ρ > 1$ , whenever $ρ_{k} > 1$ we get $∣ a_{k + 1} ∣ = (ρ_{k})^{k + 1} > 1^{k + 1} = 1$ ; so by step 3.2 there are indices $k \geq n$ with $∣ a_{k + 1} ∣ > 1$ for every $n$ .

step 3.2L3

4.3

In the case $ρ < 1$ : since $0 < t < 1$ the geometric series $\sum_{j \geq 0} t^{j}$ converges, hence so does its first tail series $\sum_{m \geq 1} t^{m}$ .

step 3.1L5

5.1

In the case $ρ < 1$ : putting $x_{j} : = ∣ a_{j + 1} ∣$ and $y_{j} : = t^{j + 1}$ for $j \in N$ , step 4.1 gives $0 \leq x_{j} \leq y_{j}$ for all $j \geq N$ , and $\sum_{j} y_{j}$ is the convergent series of step 4.3; so $\sum_{k \geq 1} ∣ a_{k} ∣$ converges.

step 4.1step 4.3L4L6

5.2

In the case $ρ > 1$ : the sequence $j \mapsto a_{j + 1}$ does not converge to $0$ , because with the rational tolerance $1$ no index $K$ satisfies $∣ a_{k + 1} ∣ < 1$ for all $k \geq K$ ; hence $\sum_{k \geq 1} a_{k}$ diverges, which is claim 2.

step 4.2L7

6.1

In the case $ρ < 1$ : the series $\sum_{k \geq 1} ∣ a_{k} ∣$ having been shown to converge, the sequence $j \mapsto a_{j + 1}$ has a convergent absolute-value series, so $\sum_{k \geq 1} a_{k}$ converges as well; together with the convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ that is claim 1.

step 5.1L11

7.1

The three cases $ρ < 1$ , $ρ > 1$ and $ρ = 1$ exhaust $\overline{R}$ , the extended order being total, so the three claims together cover every family.

step 6.1step 5.2step 3.3L1cases-exhaustive∎

Remarks

The test reads only the tail suprema, and that is why it never needs the roots to converge. Claim 1 uses a single index $N$ beyond which all roots sit below a fixed $t < 1$ ; claim 2 uses only that roots above $1$ occur arbitrarily late. Neither argument asks whether $(ρ_{k})$ has a limit, which is exactly the advantage of $lim sup$ over $\lim$ here.
Claim 2 is proved through the term test, not through a comparison. What the hypothesis delivers is infinitely many terms of absolute value greater than $1$ , which already forbids the terms from tending to $0$ . No estimate on the partial sums is needed, and none is available, the terms having no sign.
The witnesses in claim 3 are chosen so that both root computations reduce to the single standard limit $n^{1 / n} \to 1$ . The companion page carries the same phenomenon with the exponents $- 1 / 2$ and $- 2$ , where the divergent witness is not the harmonic series.

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

Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} \neq 0$ for every $k \in N$ and put

$q_{k} : = ∣ \frac{a_{k + 1}}{a_{k}} ∣ = \frac{∣ a_{k + 1} ∣}{∣ a_{k} ∣} (k \in N),$

a genuine sequence on $N$ , whose limit superior and limit inferior exist in $\overline{R}$ for every such $(a_{k})$ (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence). Then:

if ${lim sup}_{k} q_{k} < 1$ then $\sum ∣ a_{k} ∣$ converges, and hence $\sum a_{k}$ converges as well (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges);
if ${lim inf}_{k} q_{k} > 1$ then $\sum a_{k}$ diverges.

The hypothesis $a_{k} \neq 0$ is what makes $q_{k}$ exist and is not a convenience: a single vanishing term leaves the ratio at that index undefined. For a family from a starting index $m$ the statement is the one above applied to the shifted sequence $j \mapsto a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), whose ratios are $∣ a_{j + m + 1} / a_{j + m} ∣$ .

Nothing is claimed when ${lim sup}_{k} q_{k} \geq 1 \geq {lim inf}_{k} q_{k}$ . In that regime the test is silent, and it has to be: the companion page carries a convergent series whose ratios have limit superior $2$ , and both a convergent and a divergent series with ratio limit exactly $1$ .

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ ; the ratios $q_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ ; the tail bounds $s_{n} = \sup {q_{k} : k \geq n}$ and $i_{n} = \inf {q_{k} : k \geq n}$ taken in $\overline{R}$ , so that ${lim sup}_{k} q_{k} = \inf {s_{n}}$ and ${lim inf}_{k} q_{k} = \sup {i_{n}}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined); and the assumption that one of the two hypotheses of the Statement holds.

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, and the extended order is total (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ). In particular $\inf {s_{n}} \leq s_{n}$ and $\sup {i_{n}} \geq i_{n}$ for every $n$ ; $q_{k} \leq s_{n}$ and $q_{k} \geq i_{n}$ for every $k \geq n$ ; a real $t$ exceeding $\inf {s_{n}}$ is not a lower bound of ${s_{n}}$ ; and a real $u$ below $\sup {i_{n}}$ is not an upper bound of ${i_{n}}$ .

[L2]

Both quantities exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L3]

Absolute value: $∣ x ∣ \geq 0$ ; $∣ x ∣ = 0$ exactly when $x = 0$ ; $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , so $∣ a_{k + 1} / a_{k} ∣ = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ (Basic properties of the absolute value).

[L4]

The principle of induction (The principle of mathematical induction).

[L5]

The geometric series $\sum_{j \geq 0} t^{j}$ converges when $∣ t ∣ < 1$ ; a series converges if and only if each of its tail series converges; and $\sum c x_{j}$ converges when $\sum x_{j}$ does (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Convergent series add and scale termwise).

[L6]

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

[L7]

If a series converges then its terms tend to $0$ (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals); and for every real $c > 0$ there is a natural $n \geq 1$ with the rational $1 / n < c$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L8]

Powers: $t^{0} = 1$ , $t^{j + 1} = t^{j} t$ , and $t^{j} > 0$ for $t > 0$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L9]

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

Proof

technique · cases

1.1

Assume ${lim sup}_{k} q_{k} < 1$ .

assume-case lt

1.2

Assume instead ${lim inf}_{k} q_{k} > 1$ .

assume-case gt

1.3

Each $q_{k}$ is a nonnegative real, being a quotient of a nonnegative real by a positive one, so every $s_{n} \geq q_{n} \geq 0$ and hence ${lim sup}_{k} q_{k} \geq 0$ .

givenL1L2L3

2.1

In the case ${lim sup}_{k} q_{k} < 1$ the value $Λ : = {lim sup}_{k} q_{k}$ therefore lies strictly between the reals $0$ and $1$ inclusive of $0$ , so it is a real number; put $t : = (Λ + 1) / 2$ , a real with $0 \leq Λ < t < 1$ .

step 1.1step 1.3L1choose

2.2

In the case ${lim inf}_{k} q_{k} > 1$ , the real $1$ is not an upper bound of ${i_{n}}$ , so there is $N \in N$ with $i_{N} > 1$ .

step 1.2L1choose

3.1

In the case $Λ < 1$ : since $t > Λ = \inf {s_{n}}$ , the real $t$ is not a lower bound of ${s_{n}}$ , so there is $N$ with $s_{N} < t$ , and then $q_{k} \leq s_{N} < t$ for every $k \geq N$ .

step 2.1L1choose

3.2

In the case ${lim inf}_{k} q_{k} > 1$ : $q_{k} \geq i_{N} > 1$ for every $k \geq N$ , since $i_{N}$ is a lower bound of ${q_{k} : k \geq N}$ .

step 2.2L1

3.3

In the case $Λ < 1$ : the series $\sum_{j \geq 0} t^{j}$ converges since $∣ t ∣ = t < 1$ , hence so does $\sum_{j \geq 0} ∣ a_{N} ∣ t^{j}$ .

step 2.1L5

4.1

In the case $Λ < 1$ : for $k \geq N$ , $∣ a_{k + 1} ∣ / ∣ a_{k} ∣ < t$ with $∣ a_{k} ∣ > 0$ , hence $∣ a_{k + 1} ∣ < t ∣ a_{k} ∣$ .

step 3.1L3algebra

4.2

In the case ${lim inf}_{k} q_{k} > 1$ : for $k \geq N$ , $∣ a_{k + 1} ∣ > ∣ a_{k} ∣$ , again multiplying by $∣ a_{k} ∣ > 0$ .

step 3.2L3algebra

5.1

In the case $Λ < 1$ : an induction on $j$ gives $∣ a_{N + j} ∣ \leq ∣ a_{N} ∣ t^{j}$ for every $j \in N$ . At $j = 0$ this is an equality, since $t^{0} = 1$ ; and if it holds at $j$ then $∣ a_{N + j + 1} ∣ < t ∣ a_{N + j} ∣ \leq t ∣ a_{N} ∣ t^{j} = ∣ a_{N} ∣ t^{j + 1}$ , using $t > 0$ .

step 4.1L4L8

5.2

In the case ${lim inf}_{k} q_{k} > 1$ : an induction on $j$ gives $∣ a_{N + j} ∣ \geq ∣ a_{N} ∣ > 0$ for every $j \in N$ . At $j = 0$ it is an equality, and if it holds at $j$ then $∣ a_{N + j + 1} ∣ > ∣ a_{N + j} ∣ \geq ∣ a_{N} ∣$ .

step 4.2L3L4

6.1

In the case $Λ < 1$ : with $x_{j} : = ∣ a_{N + j} ∣$ and $y_{j} : = ∣ a_{N} ∣ t^{j}$ we have $0 \leq x_{j} \leq y_{j}$ for every $j$ , so $\sum_{j} ∣ a_{N + j} ∣$ converges; that is the $N$ -th tail series of $\sum ∣ a_{k} ∣$ , so $\sum ∣ a_{k} ∣$ converges.

step 5.1step 3.3L3L5L6

6.2

In the case ${lim inf}_{k} q_{k} > 1$ : $(a_{k})$ does not converge to $0$ . Choose a natural $n \geq 1$ with $1 / n < ∣ a_{N} ∣$ ; if $a_{k} \to 0$ there would be $K$ with $∣ a_{k} ∣ < 1 / n$ for all $k \geq K$ , contradicting $∣ a_{k} ∣ \geq ∣ a_{N} ∣ > 1 / n$ at any index $k$ that is at least both $K$ and $N$ .

step 5.2L7choose

7.1

In the case ${lim inf}_{k} q_{k} > 1$ : by the term test $\sum a_{k}$ diverges, which is claim 2.

step 6.2L7

7.2

In the case $Λ < 1$ : the series $\sum ∣ a_{k} ∣$ having been shown to converge, $\sum a_{k}$ converges as well; together with the convergence of $\sum ∣ a_{k} ∣$ that is claim 1.

step 6.1L9

8.1

The two assumed hypotheses are the cases of the disjunction in the Given, and they exhaust it; outside them both claims are vacuous, each hypothesis being false, so the theorem holds for every sequence with nonvanishing terms.

step 7.2step 7.1cases-exhaustive∎

Remarks

The two halves are not dual, and the asymmetry is real. Convergence needs the ratios to be eventually below a fixed $t < 1$ , which $lim sup < 1$ supplies; divergence needs them eventually above $1$ , which is what $lim inf > 1$ supplies. A hypothesis on $lim sup$ alone can never force divergence, since a single large ratio occurring arbitrarily late says nothing about the size of the terms. That is exactly what FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges records.
The geometric series is the only convergent series the proof knows. Claim 1 is a comparison against $∣ a_{N} ∣ t^{j}$ , and every later refinement on this page, Kummer's test included, exists because that comparison is too coarse when the ratios approach $1$ .

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

Whenever the ratio test decides, the root test decides the same way, and the converse fails

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} \neq 0$ for every $k \in N$ , and put

$q_{k} : = \frac{∣ a_{k + 1} ∣}{∣ a_{k} ∣}, ρ_{k} : = ∣ a_{k + 1} ∣^{1 / (k + 1)} (k \in N),$

the ratio and root families of Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence and Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing. Then, in $\overline{R}$ ,

${lim inf}_{k} q_{k} \leq {lim inf}_{k} ρ_{k} \leq {lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k},$

and consequently:

if ${lim sup}_{k} q_{k} < 1$ , so that the ratio test gives convergence of $\sum ∣ a_{k} ∣$ and hence of $\sum a_{k}$ , then ${lim sup}_{k} ρ_{k} < 1$ and the root test gives the same;
if ${lim inf}_{k} q_{k} > 1$ , so that the ratio test gives divergence of $\sum a_{k}$ , then ${lim sup}_{k} ρ_{k} > 1$ and the root test gives it too.

The converse fails. Let $(s_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , let $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ , and put $a_{k} : = 2^{- k} t_{k}$ , the sequence usually written $a_{k} = 2^{- k + (- 1)^{k}}$ . For it, ${lim sup}_{k} ρ_{k} = 1 / 2 < 1$ while ${lim sup}_{k} q_{k} = 2$ and ${lim inf}_{k} q_{k} = 1 / 8$ (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence), so the root test gives convergence of $\sum ∣ a_{k} ∣$ and neither half of the ratio test applies. So the root test decides strictly more series than the ratio test.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ , the ratios $q_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ and the roots $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L1]

For a sequence $(b_{k})$ of reals with $b_{k} > 0$ for every $k$ , writing $q_{k}^{'} = b_{k + 1} / b_{k}$ and $r_{k} = b_{k + 1}^{1 / (k + 1)}$ , one has ${lim inf}_{k} q_{k}^{'} \leq {lim inf}_{k} r_{k} \leq {lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}^{'}$ in $\overline{R}$ (For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ ).

[L2]

${lim inf}_{k} x_{k} \leq {lim sup}_{k} x_{k}$ for every real sequence ( $lim inf x_{k} \leq lim sup x_{k}$ for every real sequence).

[L3]

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

[L4]

The root test: for a family from $1$ , ${lim sup}_{k} ∣ a_{k + 1} ∣^{1 / (k + 1)} < 1$ gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ and hence of $\sum_{k \geq 1} a_{k}$ , and $> 1$ gives divergence of $\sum_{k \geq 1} a_{k}$ (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: ${lim sup}_{k} q_{k} < 1$ gives convergence of $\sum ∣ a_{k} ∣$ and hence of $\sum a_{k}$ , and ${lim inf}_{k} q_{k} > 1$ gives divergence of $\sum a_{k}$ (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L6]

A series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

For the sequence $a_{k} = 2^{- k} t_{k}$ built from the alternating sequence as in the Statement: ${lim sup}_{k} ∣ a_{k + 1} ∣^{1 / (k + 1)} = 1 / 2$ , ${lim sup}_{k} ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = 2$ and ${lim inf}_{k} ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = 1 / 8$ ; and $2^{- k} > 0$ , $t_{k} > 0$ , so every term is positive and in particular nonzero (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive 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 $s \circ o \equiv - 1$ , Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

Proof

technique · direct

1.1

Put $b_{k} : = ∣ a_{k} ∣$ . Since $a_{k} \neq 0$ we have $b_{k} > 0$ for every $k$ , so [L1] applies to $(b_{k})$ .

givenL3L1

1.2

For the sequence $a_{k} = 2^{- k} t_{k}$ of the Statement every term is nonzero, ${lim sup}_{k} ρ_{k} = 1 / 2 < 1$ , and neither ${lim sup}_{k} q_{k} = 2 < 1$ nor ${lim inf}_{k} q_{k} = 1 / 8 > 1$ holds.

2.1

For this $(b_{k})$ the ratio family is $b_{k + 1} / b_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = q_{k}$ and the root family is $b_{k + 1}^{1 / (k + 1)} = ∣ a_{k + 1} ∣^{1 / (k + 1)} = ρ_{k}$ .

step 1.1

3.1

Therefore ${lim inf}_{k} q_{k} \leq {lim inf}_{k} ρ_{k} \leq {lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k}$ , which is the displayed chain.

step 1.1step 2.1L1

4.1

Suppose ${lim sup}_{k} q_{k} < 1$ . By the chain, ${lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k} < 1$ , so the root test applies to the family $(a_{k})_{k \geq 1}$ and gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ and of $\sum_{k \geq 1} a_{k}$ , hence of $\sum ∣ a_{k} ∣$ and of $\sum a_{k}$ ; the ratio test gives the same conclusions. That is claim 1.

step 3.1L4L5L6

4.2

Suppose ${lim inf}_{k} q_{k} > 1$ . By the chain and [L2], ${lim sup}_{k} ρ_{k} \geq {lim inf}_{k} ρ_{k} \geq {lim inf}_{k} q_{k} > 1$ , so the root test gives divergence of $\sum_{k \geq 1} a_{k}$ , hence of $\sum a_{k}$ ; the ratio test gives the same conclusion. That is claim 2.

step 3.1L2L4L5L6

5.1

So for that sequence the root test gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ while neither half of the ratio test applies, and the converse of claims 1 and 2 fails.

step 1.2L4L5∎

Remarks

The dominance is a statement about $lim sup$ , not about series. The whole content is the chain of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ , proved on the previous page precisely because it is about limits superior and nothing else. Claims 1 and 2 are the translation of that chain through the two tests, and they carry no further mathematics.
Strictly more, not merely at least as much. The witness in the Statement settles that: its roots converge to $1 / 2$ while its ratios oscillate between $1 / 8$ and $2$ , so the ratio test is silent about a series the root test decides. The reason is structural rather than accidental. Taking an $n$ -th root divides the exponent by $n$ and so damps a bounded oscillation, while forming a ratio differences the exponent and preserves it.
The ratio test survives because it is easier to compute. Nothing here says the ratio test should be abandoned; the ratios of a series given by an explicit formula are usually elementary, and the roots usually are not.

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

Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges

Statement

Let $(a_{k})$ and $(ζ_{k})$ be sequences of reals with

$a_{k} > 0 and ζ_{k} > 0 for every k \in N,$

and define Kummer's expression

$K_{k} : = ζ_{k} \frac{a_{k}}{a_{k + 1}} - ζ_{k + 1} (k \in N),$

a sequence of reals whose limit inferior exists in $\overline{R}$ (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence). Then:

if ${lim inf}_{k} K_{k} > 0$ then $\sum a_{k}$ converges;
if $\sum 1 / ζ_{k}$ diverges and $K_{k} \leq 0$ for all $k$ from some index on, then $\sum a_{k}$ diverges.

Positivity of $(a_{k})$ is load bearing and is not a normalisation. Claim 2 is FALSE for terms of mixed sign, and the failure is not delicate: see the first remark below, where a convergent geometric series with negative ratio satisfies every hypothesis of claim 2 with the weights $ζ_{k} = 1$ .

The two claims specialise to the ratio test at $ζ_{k} = 1$ and to Raabe's test at $ζ_{k} = k + 1$ ; those two corollaries follow immediately below, and they are the only ways this theorem is used on this page.

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(ζ_{k})$ of reals with $a_{k} > 0$ and $ζ_{k} > 0$ for every $k$ ; Kummer's expression $K_{k} = ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}$ ; the auxiliary sequence $b_{k} : = ζ_{k} a_{k}$ , which is positive; and the tail infima $i_{n} = \inf {K_{k} : k \geq n}$ taken in $\overline{R}$ , so that ${lim inf}_{k} K_{k} = \sup {i_{n} : n \in N}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L1]

[L2]

Both limit quantities exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L3]

A nonincreasing sequence bounded below converges (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set); consecutive comparisons suffice to establish monotonicity (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L4]

$\sum (x_{k} - x_{k + 1})$ converges whenever $(x_{k})$ converges ( $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ ).

[L5]

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

[L6]

For $c \neq 0$ , $\sum c x_{k}$ converges if and only if $\sum x_{k}$ converges (Convergent series add and scale termwise); and a series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

The principle of induction (The principle of mathematical induction); and $1 / x > 0$ for $x > 0$ , with $x \leq y$ implying $1 / y \leq 1 / x$ for positive $x, y$ (Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct

1.1

Suppose ${lim inf}_{k} K_{k} > 0$ . The real $0$ is then not an upper bound of ${i_{n}}$ , so there is $N \in N$ with $i_{N} > 0$ .

givenL1L2choose

1.2

Suppose now that $\sum 1 / ζ_{k}$ diverges and that there is $N$ with $K_{k} \leq 0$ for every $k \geq N$ .

given

2.1

Since $i_{N} \leq K_{N}$ and $i_{N} > 0$ , the value $i_{N}$ is a real number; put $c : = i_{N} > 0$ , so that $K_{k} \geq c$ for every $k \geq N$ .

step 1.1L1

2.2

Multiplying $K_{k} \leq 0$ by $a_{k + 1} > 0$ gives $ζ_{k} a_{k} \leq ζ_{k + 1} a_{k + 1}$ , that is $b_{k} \leq b_{k + 1}$ , for every $k \geq N$ .

step 1.2givenalgebra

3.1

Multiplying $K_{k} \geq c$ by $a_{k + 1} > 0$ gives $ζ_{k} a_{k} - ζ_{k + 1} a_{k + 1} \geq c a_{k + 1}$ , that is $b_{k} - b_{k + 1} \geq c a_{k + 1} > 0$ , for every $k \geq N$ .

step 2.1givenalgebra

3.2

An induction on $j$ gives $b_{N + j} \geq b_{N}$ for every $j \in N$ : at $j = 0$ it is an equality, and if it holds at $j$ then $b_{N + j + 1} \geq b_{N + j} \geq b_{N}$ .

step 2.2L7

4.1

Hence $b_{N + j} \geq b_{N + j + 1}$ for every $j \in N$ , so the tail sequence $(b_{N + j})_{j}$ is nonincreasing; and it is bounded below by $0$ , every $b_{k}$ being positive.

step 3.1givenL3

4.2

So $ζ_{m} a_{m} \geq b_{N} > 0$ for every $m \geq N$ , and dividing by $ζ_{m} > 0$ gives $a_{m} \geq b_{N} \cdot (1 / ζ_{m}) > 0$ .

step 3.2givenL7

5.1

Therefore $(b_{N + j})_{j}$ converges, and by the telescoping lemma $\sum_{j} (b_{N + j} - b_{N + j + 1})$ converges.

step 4.1L3L4

5.2

Since $\sum 1 / ζ_{k}$ diverges and $b_{N} \neq 0$ , the series $\sum b_{N} (1 / ζ_{k})$ diverges.

step 1.2step 4.2L6

6.1

By step 3.1 we have $0 \leq c a_{N + j + 1} \leq b_{N + j} - b_{N + j + 1}$ for every $j$ , so $\sum_{j} c a_{N + j + 1}$ converges by comparison, and since $c \neq 0$ so does $\sum_{j} a_{N + j + 1}$ .

step 3.1step 5.1L5L6

7.1

That last series is the $(N + 1)$ -th tail series of $\sum a_{k}$ , so $\sum a_{k}$ converges, which is claim 1.

step 6.1L6

8.1

If $\sum a_{k}$ converged then, since $0 \leq b_{N} (1 / ζ_{m}) \leq a_{m}$ for $m \geq N$ , comparison would make $\sum b_{N} (1 / ζ_{k})$ converge, contradicting step 5.2; so $\sum a_{k}$ diverges, which is claim 2.

step 4.2step 5.2L5∎

Remarks

Claim 2 fails for terms of mixed sign, and here is the witness. Take $ζ_{k} = 1$ for every $k$ and $a_{k} = (- 1 / 2)^{k}$ . Then $a_{k} / a_{k + 1} = 1 / (- 1 / 2) = - 2$ (Laws of integer exponents, Integer powers $a^{m}$ ), so $K_{k} = - 2 - 1 = - 3 \leq 0$ at every index; and $\sum 1 / ζ_{k} = \sum 1$ diverges, its terms not tending to $0$ (If a series converges then its terms tend to $0$ ). Both hypotheses of claim 2 hold. Yet $\sum (- 1 / 2)^{k}$ converges, with sum $2 / 3$ , since $∣ - 1 / 2 ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges). The conclusion therefore fails, and what fails with it is exactly step 2.2, which multiplied an inequality by $a_{k + 1}$ and needed that factor to be positive. The classical signed witness at $ζ_{k} = 1$ is $a_{k} = (- 1)^{k} / k$ , whose hypotheses check the same way; its convergence is the alternating series test, which this page does not prove, and that is why the geometric witness is the one used here.
The weights are a free parameter, and that is the point of the theorem. Kummer's test is not a single criterion but a family of them, one for each positive sequence $(ζ_{k})$ , and the strength of the resulting test is exactly the strength of the divergent comparison series $\sum 1 / ζ_{k}$ it carries. Constant weights give the ratio test, weights $k + 1$ give Raabe's test, and the pattern continues past what this page can state, since the next natural choice needs the logarithm.
Claim 1 does not need $\sum 1 / ζ_{k}$ to diverge. The convergence half uses only positivity of the weights, through the telescoping bound in step 5.1. The divergence half is where the weights have to be tied to a known divergent series, and that asymmetry is why the two halves are not mirror images.

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

Kummer with $ζ_{k} = 1$ recovers the ratio test

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} > 0$ for every $k \in N$ , and put $q_{k} : = a_{k + 1} / a_{k}$ , which is the ratio family of Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence since $∣ a_{k} ∣ = a_{k}$ here. Take the constant weights $ζ_{k} : = 1$ , so that Kummer's expression (Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges) is

$K_{k} = \frac{a_{k}}{a_{k + 1}} - 1 (k \in N) .$

Then:

if ${lim sup}_{k} q_{k} < 1$ then ${lim inf}_{k} K_{k} > 0$ , so Kummer's convergence criterion applies and yields convergence of $\sum a_{k}$ ;
if ${lim inf}_{k} q_{k} > 1$ then $\sum 1 / ζ_{k}$ diverges and $K_{k} \leq 0$ from some index on, so Kummer's divergence criterion applies and yields divergence of $\sum a_{k}$ .

Both conclusions are exactly those of Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence for a sequence of positive terms. So the ratio test is the constant-weight case of Kummer's test, and every strengthening of Kummer's test by a better choice of weights is a strengthening of the ratio test.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} > 0$ for every $k$ ; the ratios $q_{k} = a_{k + 1} / a_{k}$ , which are positive; the constant weights $ζ_{k} = 1$ ; and Kummer's expression $K_{k} = a_{k} / a_{k + 1} - 1$ (Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined); ${lim sup}_{k} x_{k} = \inf {s_{n}}$ and ${lim inf}_{k} x_{k} = \sup {i_{n}}$ with $s_{n}, i_{n}$ the tail bounds, and both exist for every sequence (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L2]

Kummer's test, in both halves, for positive terms and positive weights (Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges).

[L3]

Reciprocation on the positives: $0 < x < y$ implies $0 < 1 / y < 1 / x$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

A series whose terms do not tend to $0$ diverges (If a series converges then its terms tend to $0$ ).

[L5]

For positive terms $∣ a_{k} ∣ = a_{k}$ , so the ratio family of Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence is the family $q_{k}$ above (Basic properties of the absolute value).

Proof

technique · direct

1.1

The constant weights $ζ_{k} = 1$ are positive, and the terms $a_{k}$ are positive, so Kummer's test applies with these data and its expression is $K_{k} = a_{k} / a_{k + 1} - 1$ .

givenL2

1.2

Each $q_{k}$ is positive, so ${lim sup}_{k} q_{k} \geq 0$ , every tail supremum being at least $q_{n} > 0$ .

givenL1

1.3

Suppose instead ${lim inf}_{k} q_{k} > 1$ . The real $1$ is not an upper bound of ${i_{n}^{'}}$ , the tail infima of $(q_{k})$ , so there is $N$ with $i_{N}^{'} > 1$ , and then $q_{k} \geq i_{N}^{'} > 1$ for every $k \geq N$ .

givenL1choose

1.4

The weight series $\sum 1 / ζ_{k}$ is $\sum 1$ , whose terms are constantly $1$ and so do not tend to $0$ ; it diverges.

givenL4

2.1

Suppose ${lim sup}_{k} q_{k} < 1$ . Then $Λ : = {lim sup}_{k} q_{k}$ lies between the reals $0$ and $1$ and is therefore real; put $t : = (Λ + 1) / 2$ , so that $Λ < t < 1$ and $t \geq 1 / 2 > 0$ .

step 1.2L1choose

2.2

For $k \geq N$ : $a_{k + 1} / a_{k} > 1$ gives $a_{k + 1} > a_{k} > 0$ , hence $a_{k} / a_{k + 1} < 1$ , hence $K_{k} < 0$ and in particular $K_{k} \leq 0$ .

step 1.3L3algebra

3.1

Since $t > \inf {s_{n}}$ , the real $t$ is not a lower bound of ${s_{n}}$ , so there is $N$ with $s_{N} < t$ , and then $q_{k} \leq s_{N} < t$ for every $k \geq N$ .

step 2.1L1choose

3.2

Kummer's divergence criterion therefore applies and $\sum a_{k}$ diverges, which is claim 2.

step 2.2step 1.4step 1.1L2

4.1

For $k \geq N$ : $0 < a_{k + 1} / a_{k} < t$ , so $a_{k} / a_{k + 1} > 1 / t$ , and hence $K_{k} = a_{k} / a_{k + 1} - 1 > 1 / t - 1 = : c$ , where $c > 0$ because $0 < t < 1$ gives $1 / t > 1$ .

step 3.1L3algebra

5.1

So $c$ is a lower bound of ${K_{k} : k \geq N}$ , whence $i_{N} \geq c$ and ${lim inf}_{k} K_{k} = \sup {i_{n}} \geq i_{N} \geq c > 0$ .

step 4.1L1

6.1

Kummer's convergence criterion therefore applies and $\sum a_{k}$ converges, which is claim 1.

step 5.1step 1.1L2

7.1

The hypotheses in claims 1 and 2 are precisely those of the two halves of the ratio test for this sequence, and the conclusions agree, so the ratio test for positive terms is the case $ζ_{k} = 1$ of Kummer's test.

step 6.1step 3.2L5∎

Remarks

What this corollary is for. It is not a new criterion. It fixes the place of the ratio test inside the Kummer family, so that the later choices of weights on this page can be read as improvements on a known test rather than as unrelated criteria.
The ratio test proved earlier is more general in one respect. It allows terms of either sign, provided none vanishes, and concludes convergence of $\sum ∣ a_{k} ∣$ . Kummer's test needs positivity throughout, so the identification above is between the positive-term case of the ratio test and the constant-weight case of Kummer's test, and it says nothing about signed terms.

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

Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} > 0$ for every $k \in N$ . Write $k + 1$ for the canonical natural $ι (k + 1) \in R$ , which is positive (Canonical naturals are positive and strictly increasing), take the weights $ζ_{k} : = k + 1$ in Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges, and put

$R_{k} : = (k + 1) (\frac{a_{k}}{a_{k + 1}} - 1) (k \in N),$

so that Kummer's expression for these weights is $K_{k} = R_{k} - 1$ . Then:

if ${lim inf}_{k} R_{k} > 1$ then $\sum a_{k}$ converges;
if ${lim sup}_{k} R_{k} < 1$ then $\sum a_{k}$ diverges.

The weights are $k + 1$ rather than $k$ because $ζ_{0}$ has to be positive and $N$ contains $0$ ; the classical statement, indexed from $1$ , is the same criterion read along the shift $k = j + 1$ .

Nothing is claimed when ${lim inf}_{k} R_{k} \leq 1 \leq {lim sup}_{k} R_{k}$ . The Gauss test proved next is exactly the tool for the borderline case $R_{k} \to 1$ , where Raabe's test is silent.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} > 0$ for every $k$ ; the weights $ζ_{k} = ι (k + 1)$ ; and $R_{k} = (k + 1) (a_{k} / a_{k + 1} - 1)$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , Canonical naturals are positive and strictly increasing).

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, and ${lim sup}_{k} x_{k} = \inf {s_{n}}$ , ${lim inf}_{k} x_{k} = \sup {i_{n}}$ for the tail bounds $s_{n}, i_{n}$ , both existing for every sequence (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L2]

Kummer's test in both halves, for positive terms and positive weights (Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges).

[L3]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; at $p = 1$ it therefore diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). Moreover $k^{1} = ι (k)$ , the rational power at exponent $1$ being the element itself (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 $a^{m}$ ).

[L4]

The canonical naturals $ι (k + 1)$ are positive, and $ι (k + 2) = ι (k + 1) + 1$ (Canonical naturals are positive and strictly increasing).

[L5]

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

Proof

technique · direct

1.1

The weights $ζ_{k} = ι (k + 1)$ are positive for every $k \in N$ , and the terms $a_{k}$ are positive, so Kummer's test applies with these data.

givenL4L2

1.2

Suppose ${lim inf}_{k} R_{k} > 1$ . The real $1$ is not an upper bound of the set of tail infima of $(R_{k})$ , so there is $N$ with $i_{N} > 1$ , and $i_{N}$ is real because $i_{N} \leq R_{N}$ .

givenL1choose

1.3

Suppose instead ${lim sup}_{k} R_{k} < 1$ . The real $1$ is not a lower bound of the set of tail suprema of $(R_{k})$ , so there is $N$ with $s_{N} < 1$ , and then $R_{k} \leq s_{N} < 1$ for every $k \geq N$ .

givenL1choose

2.1

Kummer's expression for these weights is $K_{k} = (k + 1) \frac{a_{k}}{a_{k + 1}} - (k + 2) = (k + 1) (\frac{a_{k}}{a_{k + 1}} - 1) - 1 = R_{k} - 1$ .

step 1.1L4algebra

2.2

The weight series is $\sum_{k} 1 / ζ_{k} = \sum_{k} 1 / ι (k + 1)$ , which is precisely the series $\sum_{k \geq 1} 1 / k$ from the starting index $1$ , and that is the case $p = 1$ of the $p$ -series, hence divergent.

step 1.1L3L5

3.1

Put $c : = i_{N} - 1 > 0$ . For every $k \geq N$ we have $R_{k} \geq i_{N}$ , hence $K_{k} = R_{k} - 1 \geq i_{N} - 1 = c$ .

step 1.2step 2.1L1algebra

3.2

Hence $K_{k} = R_{k} - 1 < 0$ , in particular $K_{k} \leq 0$ , for every $k \geq N$ .

step 1.3step 2.1algebra

4.1

So $c$ is a lower bound of ${K_{k} : k \geq N}$ , whence ${lim inf}_{k} K_{k} \geq c > 0$ , and Kummer's convergence criterion gives convergence of $\sum a_{k}$ , which is claim 1.

step 3.1step 1.1L1L2

5.1

Together with the divergence of the weight series, Kummer's divergence criterion gives divergence of $\sum a_{k}$ , which is claim 2.

step 3.2step 2.2step 1.1L2∎

Remarks

Raabe's test is a genuine strengthening of the ratio test. Whenever the ratios $a_{k + 1} / a_{k}$ converge to $1$ the ratio test is silent, while $R_{k}$ may still be bounded away from $1$ on either side; the companion page carries a series with ratio limit exactly $1$ that Raabe decides. The reason is visible in the weights: the divergent comparison series behind the test has moved from $\sum 1$ to the harmonic series, which diverges far more slowly.
The threshold is $1$ and not $0$ , and step 2.1 says why. Kummer's criterion is a statement about $K_{k} = R_{k} - 1$ ; the shift by $1$ between the two expressions is the whole difference between the two thresholds, and it comes from $ζ_{k + 1} - ζ_{k} = 1$ for these weights.

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

Gauss: for positive terms, if $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ , some constant $C$ and some rational $ε > 0$ , the series converges iff $h > 1$

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} > 0$ for every $k \in N$ . Suppose there are a real $h$ , a real $C \geq 0$ , a rational $ε > 0$ and reals $r_{k}$ for $k \geq 1$ such that

$\frac{a_{k}}{a_{k + 1}} = 1 + \frac{h}{k} + r_{k} and ∣ r_{k} ∣ \leq C k^{- 1 - ε} (k \geq 1),$

where $k$ denotes the canonical natural $ι (k) > 0$ and $k^{- 1 - ε}$ is the rational power (Rational powers $a^{r}$ of a positive base, Canonical naturals are positive and strictly increasing). Then

$\sum a_{k} converges ⟺ h > 1.$

The hypotheses are imposed from $k = 1$ on, since $h / k$ has no value at $k = 0$ ; $a_{0}$ is unconstrained beyond being positive, which costs nothing because convergence is a tail property (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

The exponent $ε$ is rational because that is what Rational powers $a^{r}$ of a positive base supplies, and the error bound is a $p$ -series bound with $p = 1 + ε > 1$ , which is exactly the summability the proof consumes.

The borderline case $h = 1$ is the whole point of the theorem. There $R_{k} = (k + 1) (a_{k} / a_{k + 1} - 1)$ tends to $1$ , so both halves of Raabe's test (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence) are silent; the theorem asserts divergence there, and the argument below establishes it without any logarithm, by a telescoping product estimate.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} > 0$ for every $k$ ; reals $h$ , $C \geq 0$ , a rational $ε > 0$ and reals $r_{k}$ ( $k \geq 1$ ) with $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ and $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ ; and $R_{k} : = (k + 1) (a_{k} / a_{k + 1} - 1)$ for $k \in N$ (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

[L1]

Raabe's test: for positive terms, ${lim inf}_{k} R_{k} > 1$ gives convergence of $\sum a_{k}$ and ${lim sup}_{k} R_{k} < 1$ gives divergence (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

[L2]

Every subset of $\overline{R}$ has least upper and greatest lower bounds there; ${lim inf}_{k} x_{k} = \sup {i_{n}}$ and ${lim sup}_{k} x_{k} = \inf {s_{n}}$ for the tail bounds, both existing for every sequence (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L3]

For every real $c > 0$ there is a natural $n \geq 1$ with $1 / n < c$ , and for every real $x$ there is a natural $n$ with $ι (n) > x$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L4]

Rational powers on a positive base: $a^{r + s} = a^{r} a^{s}$ , $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ , $a^{r} > 0$ ; and for rational $t > 0$ , $0 < x < y$ implies $x^{t} < y^{t}$ (Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Rational powers $a^{r}$ of a positive base).

[L5]

Limit rules: sums, scalar multiples, products and quotients of convergent sequences (Algebra of limits: sums, scalar multiples, products and quotients); the squeeze theorem (The squeeze theorem); convergence depends only on the tail (Convergence depends only on the tail); a convergent sequence satisfies its estimate for every real tolerance (Limits and Cauchy sequences of reals, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[L7]

Direct comparison, and the fact that scaling by a nonzero constant preserves convergence and divergence (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); a series converges if and only if each of its tail series converges, and for nonnegative terms every partial sum is at most the sum (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

The principle of induction (The principle of mathematical induction); reciprocation reverses order on the positives (Inverses of positives are positive, and reciprocation reverses order); $∣ x ∣ \geq 0$ and $x \leq ∣ x ∣$ (Basic properties of the absolute value).

Proof

technique · cases

1.1

Assume $h > 1$ .

assume-case gt

1.2

Assume instead $h < 1$ .

assume-case lt

1.3

Assume instead $h = 1$ .

assume-case eq

1.4

For every $k \geq 1$ , $R_{k} = (k + 1) (\frac{h}{k} + r_{k}) = h (1 + \frac{1}{k}) + (k + 1) r_{k}$ .

givenalgebra

1.5

The sequence $j \mapsto 1 / (j + 1)$ converges to $0$ : given a rational $η > 0$ , choose a natural $n \geq 1$ with $1 / n < η$ ; then $1 / (j + 1) \leq 1 / n < η$ for every $j$ with $j + 1 \geq n$ .

L3L8choose

1.6

The sequence $j \mapsto (j + 1)^{- ε}$ converges to $0$ : given a real $η > 0$ , put $M : = \max {1 / η, 1} > 0$ and choose a natural $n$ with $ι (n) > M^{1 / ε}$ ; then for $j + 1 \geq n$ we get $(j + 1)^{ε} > (M^{1 / ε})^{ε} = M \geq 1 / η$ , hence $0 < (j + 1)^{- ε} < η$ .

L3L4L8choose

1.7

For every $j \in N$ , $∣ (j + 2) r_{j + 1} ∣ \leq C (j + 2) (j + 1)^{- 1 - ε} \leq 2 C (j + 1) (j + 1)^{- 1 - ε} = 2 C (j + 1)^{- ε}$ , using $j + 2 \leq 2 (j + 1)$ .

givenL4L8algebra

2.1

Hence $- 2 C (j + 1)^{- ε} \leq (j + 2) r_{j + 1} \leq 2 C (j + 1)^{- ε}$ with both bounds converging to $0$ , so $(j + 2) r_{j + 1} \to 0$ by the squeeze theorem.

step 1.7step 1.6L5

2.2

In the case $h = 1$ , put $u_{k} : = \frac{k}{k + 1} r_{k}$ and $t_{k} : = k a_{k}$ for $k \geq 1$ ; then $t_{k} > 0$ , $∣ u_{k} ∣ \leq ∣ r_{k} ∣ \leq C k^{- 1 - ε}$ since $0 < k / (k + 1) < 1$ , and $\frac{a_{k}}{a_{k + 1}} = \frac{k + 1}{k} + r_{k} = \frac{k + 1}{k} (1 + u_{k})$ .

step 1.3givenL8algebra

3.1

Therefore $R_{j + 1} = h (1 + 1 / (j + 1)) + (j + 2) r_{j + 1} \to h (1 + 0) + 0 = h$ , and since convergence depends only on the tail, the sequence $(R_{k})_{k \in N}$ converges to $h$ .

step 1.4step 1.5step 2.1L5

3.2

Consequently $\frac{t_{k}}{t_{k + 1}} = \frac{k}{k + 1} \cdot \frac{a_{k}}{a_{k + 1}} = 1 + u_{k}$ for $k \geq 1$ , so $1 + u_{k} = t_{k} / t_{k + 1} > 0$ and $\frac{t_{k + 1}}{t_{k}} = \frac{1}{1 + u_{k}}$ .

step 2.2algebra

3.3

In the case $h = 1$ : $\sum_{k \geq 1} k^{- 1 - ε}$ converges, since $1 + ε$ is a rational exceeding $1$ ; hence so does $\sum_{k \geq 1} C k^{- 1 - ε}$ , and by comparison with it so does $\sum_{k \geq 1} ∣ u_{k} ∣$ , whose terms are nonnegative.

step 2.2L6L7

4.1

In the case $h > 1$ : applying the limit estimate with the real tolerance $(h - 1) / 2 > 0$ gives an $N$ with $R_{k} > h - (h - 1) / 2 = (h + 1) / 2$ for all $k \geq N$ ; so $(h + 1) / 2$ is a lower bound of ${R_{k} : k \geq N}$ , whence ${lim inf}_{k} R_{k} \geq (h + 1) / 2 > 1$ and $\sum a_{k}$ converges.

step 1.1step 3.1L2L5L1

4.2

In the case $h < 1$ : the tolerance $(1 - h) / 2 > 0$ gives an $N$ with $R_{k} < h + (1 - h) / 2 = (h + 1) / 2$ for all $k \geq N$ ; so $(h + 1) / 2$ is an upper bound of ${R_{k} : k \geq N}$ , whence ${lim sup}_{k} R_{k} \leq (h + 1) / 2 < 1$ and $\sum a_{k}$ diverges.

step 1.2step 3.1L2L5L1

4.3

For $k \geq 1$ : $(1 - u_{k}) (1 + u_{k}) = 1 - u_{k}^{2} \leq 1$ and $1 + u_{k} > 0$ , so $\frac{1}{1 + u_{k}} \geq 1 - u_{k} \geq 1 - ∣ u_{k} ∣$ .

step 3.2L8algebra

4.4

Writing $U$ for the sum of $\sum_{k \geq 1} ∣ u_{k} ∣$ and $P_{n} = \sum_{k = 1}^{n} ∣ u_{k} ∣$ for its partial sums, $P_{n} \to U$ , so there is a natural $N_{0}$ with $U - P_{N_{0}} \leq 1 / 2$ ; put $N : = N_{0} + 1$ .

step 3.3L5L7choose

5.1

For every $n \geq N$ the block $\sum_{k = N}^{n} ∣ u_{k} ∣$ is a partial sum of the $N_{0}$ -th tail series of $\sum_{k \geq 1} ∣ u_{k} ∣$ , whose terms are nonnegative and whose sum is $U - P_{N_{0}}$ ; hence $\sum_{k = N}^{n} ∣ u_{k} ∣ \leq 1 / 2$ , and in particular $∣ u_{k} ∣ \leq 1 / 2$ for every $k \geq N$ .

step 4.4L7

6.1

In the case $h = 1$ : for every $n \geq N - 1$ , $\frac{t_{n + 1}}{t_{N}} \geq 1 - \sum_{k = N}^{n} ∣ u_{k} ∣$ , by induction on $n$ . At $n = N - 1$ both sides equal $1$ , the sum being empty. Assume it at $n$ ; then, since $t_{n + 1} / t_{N} > 0$ and $1 / (1 + u_{n + 1}) \geq 1 - ∣ u_{n + 1} ∣ \geq 1 / 2 > 0$ , and since the induction hypothesis gives $t_{n + 1} / t_{N} \geq 1 - \sum_{k = N}^{n} ∣ u_{k} ∣ \geq 1 / 2 > 0$ , we get $\frac{t_{n + 2}}{t_{N}} = \frac{1}{1 + u_{n + 1}} \cdot \frac{t_{n + 1}}{t_{N}} \geq (1 - ∣ u_{n + 1} ∣) (1 - \sum_{k = N}^{n} ∣ u_{k} ∣) \geq 1 - \sum_{k = N}^{n + 1} ∣ u_{k} ∣$ , the last step expanding the product and discarding a nonnegative term.

step 3.2step 4.3step 5.1L8

7.1

Hence $t_{n + 1} / t_{N} \geq 1 - 1 / 2 = 1 / 2$ for every $n \geq N - 1$ , that is $m a_{m} = t_{m} \geq t_{N} / 2 > 0$ for every $m \geq N$ , and so $a_{m} \geq \frac{t_{N}}{2} \cdot \frac{1}{m}$ for every $m \geq N$ .

step 6.1step 5.1L8algebra

8.1

The series $\sum_{m \geq 1} 1 / m$ diverges, so $\sum_{m \geq 1} \frac{t_{N}}{2} \cdot \frac{1}{m}$ diverges, the factor $t_{N} / 2$ being nonzero.

step 7.1L6L7

9.1

If $\sum a_{k}$ converged, then so would $\sum_{m \geq 1} a_{m}$ , and comparison with the estimate of step 7.1 would make $\sum_{m \geq 1} \frac{t_{N}}{2 m}$ converge, contradicting step 8.1; so in the case $h = 1$ the series $\sum a_{k}$ diverges.

step 7.1step 8.1L7

10.1

The three cases $h > 1$ , $h < 1$ and $h = 1$ exhaust the reals, and they give convergence, divergence and divergence respectively; so $\sum a_{k}$ converges exactly when $h > 1$ .

step 4.1step 4.2step 9.1cases-exhaustive∎

Remarks

No logarithm anywhere, and that is deliberate. The classical treatment of $h = 1$ compares $a_{k}$ with $1 / (k \log k)$ or invokes Bertrand's test. Neither is available in this library at this point, and neither is needed: the hypothesis $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ makes $\sum ∣ u_{k} ∣$ convergent, and a convergent sum of nonnegative errors is exactly what the product estimate of step 6.1 consumes. The price is that the theorem is stated with an $ε$ of decay to spare, rather than for an arbitrary summable error.
Step 7.1 is the Weierstrass product inequality in disguise. In the form $\prod_{j} (1 - x_{j}) \geq 1 - \sum_{j} x_{j}$ for $x_{j} \in [0, 1]$ , it is the standard statement; here the product is $t_{n + 1} / t_{N}$ , telescoped in advance, so that one induction does the work of two and no separate lemma about products of inequalities is needed.
What the conclusion at $h = 1$ says about the terms. The estimate $a_{m} \geq (t_{N} / 2) (1 / m)$ is a genuine lower bound of harmonic type: at the borderline the terms cannot decay faster than a constant multiple of $1 / m$ , and divergence follows from the divergence of the harmonic series alone.
The three cases are decided by $h$ and by nothing else. The constants $C$ and $ε$ never appear in the conclusion; they enter only through the requirement that the error be summable, which is what keeps the case $h = 1$ from being genuinely borderline in this argument.

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

For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} > 0$ for every $k \in N$ and suppose $\sum a_{k}$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series). Write

$S_{n} : = \sum_{k = 0}^{n} a_{k} (n \in N)$

for the inclusive partial sums, so that $S_{n} = s_{n + 1}$ in the notation $s_{n} = \sum_{k < n} a_{k}$ of Series, partial sums, convergence and the sum, divergence, and the tail series. Then $S_{n} > 0$ for every $n \in N$ , and:

$\sum_{n} \frac{a_{n}}{S_{n}}$ diverges;
$\sum_{n} \frac{a_{n}}{S_{n}^{2}}$ converges.

Why the divisor is the inclusive partial sum. The exclusive partial sum of Series, partial sums, convergence and the sum, divergence, and the tail series has $s_{0} = 0$ , the empty sum, so $a_{0} / s_{0}$ has no value and a series divided by $s_{n}$ would have to begin at $n = 1$ . The inclusive sum has $S_{0} = a_{0} > 0$ , so both series above are series of sequences on $N$ with no shift and no excluded index. The classical statement, which writes $s_{n} = a_{1} + \dots + a_{n}$ and starts at $n = 1$ , is this one with the indices moved by one.

What the theorem says. No divergent series of positive terms is slowest: dividing its terms by the running total produces a series that still diverges but whose terms are eventually strictly smaller. Dividing by the square of the running total overshoots and produces a convergent series. Claim 1 is what refutes the existence of a universal comparison series (FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test).

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} > 0$ for every $k$ , with $\sum a_{k}$ divergent; the exclusive partial sums $s_{n} = \sum_{k < n} a_{k}$ and the inclusive partial sums $S_{n} = \sum_{k = 0}^{n} a_{k} = s_{n + 1}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

For a series of nonnegative terms: the partial sums are nondecreasing, the series converges if and only if their range is bounded above, and otherwise they diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, A nondecreasing sequence that is not bounded above diverges to $+ \infty$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Divergence to $+ \infty$ and to $- \infty$ ).

[L2]

Splitting of finite sums: for $m \leq n$ , $\sum_{k = 0}^{n} a_{k} = \sum_{k = 0}^{m} a_{k} + \sum_{k = m + 1}^{n} a_{k}$ , so $\sum_{k = m + 1}^{n} a_{k} = S_{n} - S_{m}$ ; and monotonicity of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L3]

The Cauchy criterion: $\sum d_{k}$ converges if and only if for every real $ε > 0$ there is $N$ with $∣ \sum_{k = m + 1}^{n} d_{k} ∣ < ε$ for all $n > m \geq N$ (A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ ).

[L4]

For positive terms, $x_{n} \to 0$ if and only if $1 / x_{n} \to + \infty$ (For positive terms, null and divergence to $+ \infty$ are reciprocal, Limits and Cauchy sequences of reals).

[L5]

$\sum (b_{n} - b_{n + 1})$ converges whenever $(b_{n})$ converges ( $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ ).

[L6]

Direct comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ), and a series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L7]

Reciprocation on the positives: $0 < x \leq y$ implies $0 < 1 / y \leq 1 / x$ (Inverses of positives are positive, and reciprocation reverses order); and $x^{2} = x \cdot x$ (Integer powers $a^{m}$ ).

Proof

technique · direct

1.1

Every $S_{n}$ is a sum of positive terms, so $S_{n} > 0$ ; and $S_{n} \leq S_{n + 1}$ since $S_{n + 1} - S_{n} = a_{n + 1} > 0$ , so $(S_{n})$ is nondecreasing and $S_{m} \leq S_{n}$ whenever $m \leq n$ .

givenL2L1

1.2

Since $\sum a_{k}$ diverges and its terms are nonnegative, the exclusive partial sums are unbounded above and $s_{n} \to + \infty$ ; hence $S_{n} = s_{n + 1} \to + \infty$ , because for a given real $M$ any index bound $K$ working for $(s_{n})$ also works for $(S_{n})$ .

givenL1algebra

2.1

For all naturals $n > m$ , using $S_{k} \leq S_{n}$ for $k \leq n$ and $a_{k} > 0$ : $\sum_{k = m + 1}^{n} \frac{a_{k}}{S_{k}} \geq \sum_{k = m + 1}^{n} \frac{a_{k}}{S_{n}} = \frac{S_{n} - S_{m}}{S_{n}} = 1 - \frac{S_{m}}{S_{n}}$ .

step 1.1L2L7

2.2

Put $b_{n} : = 1 / S_{n}$ , which is positive; since $1 / b_{n} = S_{n} \to + \infty$ , the sequence $(b_{n})$ converges to $0$ .

step 1.1step 1.2L4L7

2.3

For every $n \in N$ : $b_{n} - b_{n + 1} = \frac{1}{S_{n}} - \frac{1}{S_{n + 1}} = \frac{S_{n + 1} - S_{n}}{S_{n} S_{n + 1}} = \frac{a_{n + 1}}{S_{n} S_{n + 1}} \geq \frac{a_{n + 1}}{S_{n + 1}^{2}} > 0$ , the inequality because $0 < S_{n} S_{n + 1} \leq S_{n + 1}^{2}$ .

step 1.1L2L7algebra

3.1

Let $N \in N$ be arbitrary and put $m : = N$ . Since $S_{n} \to + \infty$ there is $n > m$ with $S_{n} > 2 S_{m}$ , and then $S_{m} / S_{n} < 1 / 2$ , so the block of step 2.1 satisfies $\sum_{k = m + 1}^{n} a_{k} / S_{k} > 1 / 2$ .

step 1.2step 2.1L7choose

3.2

Therefore $\sum_{n} (b_{n} - b_{n + 1})$ converges.

step 2.2L5

4.1

So no $N$ witnesses the Cauchy condition for the tolerance $ε = 1 / 2$ , and $\sum_{n} a_{n} / S_{n}$ diverges, which is claim 1.

step 3.1L3

5.1

By comparison, $\sum_{n} \frac{a_{n + 1}}{S_{n + 1}^{2}}$ converges; that series is the $1$ -st tail series of $\sum_{n} \frac{a_{n}}{S_{n}^{2}}$ , so the latter converges, which is claim 2.

step 3.2step 2.3L6∎

Remarks

The two claims are not two theorems but one pair of estimates. Divergence comes from bounding a block below by $1 - S_{m} / S_{n}$ , which the Cauchy criterion turns into a refutation of convergence; convergence comes from bounding a single term above by a telescoping difference. The first estimate needs $S_{n}$ to grow without bound and the second needs it to be nondecreasing, and both facts come from divergence of $\sum a_{k}$ together with positivity of its terms.
The exponent $2$ is not optimal, and this page does not pursue that. The classical refinement replaces $S_{n}^{2}$ by $S_{n}^{1 + δ}$ for a rational $δ > 0$ ; the argument is the same in outline but needs an estimate for $S_{n}^{- δ} - S_{n + 1}^{- δ}$ that the tools on this page do not supply cleanly. The square is what claim 1 needs a companion for, and it is enough for every use made of the theorem here.
Positivity is used at every step. It gives $S_{n} > 0$ , so the quotients exist; it makes $(S_{n})$ nondecreasing, which both estimates use; and it makes the terms of the two derived series nonnegative, which is what lets comparison and the boundedness criterion apply to them.

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

How the nonnegative tests are ordered by strength, and which of them this page cannot state without the logarithm

The tests on this page are not independent criteria of comparable status. Some of them are strictly stronger than others, in the precise sense that whenever the weaker one decides a series, the stronger one decides it the same way, and there are series the stronger one decides and the weaker one does not. This remark records exactly which comparisons are proved here, and, equally importantly, which are not.

Everything on this page is a comparison in disguise. 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 $\sum b_{k}$ compares against an arbitrary series; the strength of every later test is the strength of the particular series it compares against. The root and ratio tests compare against a geometric series (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges); Raabe's test compares against the harmonic series, through the weights $ζ_{k} = k + 1$ in Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges; and the borderline branch of Gauss: for positive terms, if $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ , some constant $C$ and some rational $ε > 0$ , the series converges iff $h > 1$ compares against the harmonic series again. For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges is of a different kind: it does not compare, it reindexes, and that is why it settles the whole $p$ -series family (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ) in one step.

The comparisons proved on this page.

Root over ratio. Whenever the ratio test decides, the root test decides the same way, and the converse fails: whenever the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence) decides, the root test (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing) decides the same way, and there is a series the root test decides and the ratio test does not. This is a consequence of the inequality chain $lim inf q_{k} \leq lim inf ρ_{k} \leq lim sup ρ_{k} \leq lim sup q_{k}$ , proved on the previous page, and of nothing else.
Ratio as a case of Kummer. Kummer with $ζ_{k} = 1$ recovers the ratio test: for positive terms, the ratio test is Kummer's test with the constant weights $ζ_{k} = 1$ .
Raabe as a case of Kummer. Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence: Raabe's test is Kummer's test with the weights $ζ_{k} = k + 1$ .
Gauss over Raabe, inside Gauss's hypothesis. Gauss: for positive terms, if $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ , some constant $C$ and some rational $ε > 0$ , the series converges iff $h > 1$ assumes an expansion $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with a summable error. Under that hypothesis $R_{k} = (k + 1) (a_{k} / a_{k + 1} - 1)$ converges to $h$ . When $h \neq 1$ , Raabe's test already decides, and Gauss's proof says so by invoking it. When $h = 1$ Raabe's test cannot decide: ${lim inf}_{k} R_{k} > 1$ would force $R_{k}$ to stay above a fixed number greater than $1$ from some index on, and ${lim sup}_{k} R_{k} < 1$ would force it to stay below a fixed number less than $1$ , and $R_{k} \to 1$ rules out both. Gauss decides that case, and it is the reason the theorem exists.

Two comparisons that are not claimed here. Raabe's test is not compared with the root test on this page, in either direction, and nothing above should be read as ordering them. Nor is Kummer's test claimed to be universal: the choice of weights is free, and the question of which series some choice of weights decides is not addressed.

And one that is refuted. No comparison test can be final. For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges turns any divergent series of positive terms into a divergent series of positive terms with eventually smaller terms, and FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test draws the conclusion: there is no slowest divergent series of positive terms, hence no universal comparison test. The hierarchy above is therefore an initial segment of something with no last term, not an approach to a best test.

What this page cannot state, and why. Every gap below is a missing definition, not a missing proof.

The $p$ -series at irrational exponents. Rational powers $a^{r}$ of a positive base defines $a^{r}$ for rational $r$ and positive $a$ . So $\sum 1 / k^{p}$ is a well-formed expression here only for rational $p$ , and For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ is the full theorem for every exponent this page can name. Real exponents wait for the exponential and the logarithm.
Bertrand's test. Its criterion is a condition on $\log k (k (a_{k} / a_{k + 1} - 1) - 1)$ , and it is the natural next member of the Kummer family, with weights $ζ_{k} = k \log k$ . Both the weights and the criterion mention the logarithm, so neither can be written down here.
The integral test. It compares $\sum f (k)$ with $\int f$ , and the Riemann integral is developed much later in this library. Condensation is the substitute used on this page, and for the $p$ -series it does the same work.
The general form of Gauss's test. The classical statement assumes $r_{k} = O (k^{- β})$ for some real $β > 1$ . The version proved here writes $β = 1 + ε$ with $ε$ a positive rational. This loses no case covered by the classical hypothesis: given $β > 1$ , choose a rational $0 < ε < β - 1$ and weaken the eventual bound. An error of order $1 / (k \log k)$ is not a Gauss remainder at $h = 1$ ; it is the next Bertrand borderline.

A limitation that has been removed, and one that has not. The comparison with a geometric series inside Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing and Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence delivers convergence of $\sum ∣ a_{k} ∣$ and not, on its own, of $\sum a_{k}$ . That the second follows from the first is If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, proved on this page from A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ and the triangle inequality for finite sums, so both tests do reach their standard conclusion here. What is not on this page is the rest of that theory: the converse fails, and the alternating harmonic series that witnesses the failure needs the alternating series test, which is not proved here; rearrangement, the Riemann series theorem and products of series belong with it on the page that follows. Nothing above asserts a converse or identifies any sum.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges

Statement

False claim: for every sequence $(a_{k})$ of reals, if $(a_{k})$ converges to $0$ (Limits and Cauchy sequences of reals) then $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series).

What is true is the converse implication, If a series converges then its terms tend to $0$ : a convergent series has terms tending to $0$ . The claim above reverses it, and the reversal fails at the very first place one looks, the harmonic series.

The witness is $a_{k} : = 1 / (k + 1)$ for $k \in N$ , which is the family $1 / k$ , $k \geq 1$ , written as a sequence on $N$ ; by Series, partial sums, convergence and the sum, divergence, and the tail series the series of this sequence is exactly $\sum_{k \geq 1} 1 / k$ .

Facts & Assumptions

Given: The sequence $a_{k} : = 1 / ι (k + 1)$ , $k \in N$ , where $ι (k + 1)$ is the canonical natural, positive for every $k$ (Canonical naturals are positive and strictly increasing).

[L1]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ); and $0 < x \leq y$ implies $0 < 1 / y \leq 1 / x$ (Inverses of positives are positive, and reciprocation reverses order).

[L2]

Convergence to $0$ means: for every rational $ε > 0$ there is $K$ with $∣ a_{k} ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[L3]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; and $k^{1} = ι (k)$ , the rational power at exponent $1$ being the element itself (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 $a^{m}$ ).

[L4]

[L5]

The refuted claim: for every sequence of reals converging to $0$ , the associated series converges.

Refutation

technique · direct

1.1

Every term $a_{k} = 1 / ι (k + 1)$ is a positive real, the canonical naturals $ι (k + 1)$ being positive.

givenL1

1.2

The series of $(a_{k})$ is $\sum_{k \geq 1} 1 / k$ , the series from starting index $1$ of the family $1 / k$ , since that series is by definition the series of $j \mapsto 1 / ι (j + 1)$ .

givenL4

2.1

The sequence $(a_{k})$ converges to $0$ : given a rational $ε > 0$ , choose a natural $n \geq 1$ with $1 / n < ε$ ; then for every $k \geq n$ we have $ι (k + 1) \geq ι (n)$ , hence $∣ a_{k} ∣ = 1 / ι (k + 1) \leq 1 / n < ε$ .

step 1.1L1L2choose

2.2

That series is the case $p = 1$ of the $p$ -series, and $p = 1$ does not exceed $1$ , so it diverges.

step 1.2L3

3.1

So $(a_{k})$ converges to $0$ while $\sum a_{k}$ diverges, and the claim fails for this sequence.

step 2.1step 2.2L5

4.1

The claim is therefore false, and what survives of it is only the converse implication, that a convergent series has null terms.

step 3.1L5∎

Remarks

The failure is not marginal. The harmonic series has terms tending to $0$ and partial sums diverging to $+ \infty$ , so no weakening of the false claim to "the partial sums are bounded" would rescue it either. The rate at which the terms tend to $0$ is what decides convergence, and the term test reads no rate at all.
The other tests use rate information that the term test ignores. The $p$ -series theorem distinguishes $1 / k$ from $1 / k^{2}$ . The basic root and ratio tests do not: for both sequences their relevant limit is the boundary value $1$ , so those two tests are inconclusive.

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

FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges

Statement

False claim: for every sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ , if

${lim sup}_{k} ∣ \frac{a_{k + 1}}{a_{k}} ∣ \geq 1$

then $\sum a_{k}$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

The true divergence half of the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence) has the hypothesis ${lim inf}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ , on the limit inferior and with a strict inequality. The claim above weakens it in both respects at once, and either weakening alone already destroys it.

The witness is built from the alternating sequence: with $(s_{k})$ as in 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 $s \circ o \equiv - 1$ , put $c_{k} : = 2 + s_{k}$ , so $c_{k}$ is $3$ at the even indices and $1$ at the odd ones, and let

$a_{k} : = c_{k} 2^{- k} (k \in N) .$

Its ratios take only the two values $3 / 2$ and $1 / 6$ , so their limit superior is at least $1$ , while the series converges by comparison with a geometric series.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ and the index maps $e, o$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; $c_{k} : = 2 + s_{k}$ ; and $a_{k} : = c_{k} 2^{- k}$ for $k \in N$ .

[L1]

The alternating sequence: $s_{0} = 1$ , $s_{k + 1} = - s_{k}$ , $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ , with $e$ and $o$ 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 $s \circ o \equiv - 1$ ); a strictly increasing index map satisfies $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Powers of $2$ : $2^{- k} > 0$ , $2^{- (k + 1)} = 2^{- k} / 2$ , and $2^{- k} = 1 / 2^{k}$ (Integer powers $a^{m}$ , Laws of integer exponents, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L3]

Limit superior in $\overline{R}$ : ${lim sup}_{k} x_{k} = \inf {s_{n}}$ with $s_{n} = \sup {x_{k} : k \geq n}$ ; every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there; and both quantities exist for every sequence (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L4]

The geometric series $\sum_{k} r^{k}$ converges for $∣ r ∣ < 1$ ; scaling by a nonzero constant preserves convergence; and direct comparison applies to nonnegative terms (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Convergent series add and scale termwise, 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 $\sum b_{k}$ , Basic properties of the absolute value).

[L5]

The ratio test: ${lim inf}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence, and that is the only divergence criterion it supplies (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L6]

The refuted claim: for every sequence of nonzero reals with ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ \geq 1$ , the series $\sum a_{k}$ diverges.

Refutation

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ , since $∣ s_{k} ∣ = 1$ ; so $c_{k} = 2 + s_{k}$ is $3$ or $1$ , in either case $1 \leq c_{k} \leq 3$ and $c_{k} > 0$ .

givenL1

2.1

Hence $a_{k} = c_{k} 2^{- k} > 0$ for every $k$ , so in particular $a_{k} \neq 0$ and the claim applies to $(a_{k})$ .

step 1.1L2L6

2.2

For every $k$ , $0 < a_{k} = c_{k} 2^{- k} \leq 3 \cdot 2^{- k}$ .

step 1.1L2

3.1

The ratios are $q_{k} : = ∣ a_{k + 1} / a_{k} ∣ = a_{k + 1} / a_{k} = \frac{c_{k + 1}}{c_{k}} \cdot \frac{1}{2}$ , so $q_{k} = (1 / 3) (1 / 2) = 1 / 6$ when $s_{k} = 1$ and $q_{k} = 3 \cdot (1 / 2) = 3 / 2$ when $s_{k} = - 1$ , using $s_{k + 1} = - s_{k}$ .

step 1.1step 2.1L1L2algebra

3.2

The geometric series $\sum_{k} (1 / 2)^{k}$ converges, since $∣ 1 / 2 ∣ < 1$ ; hence so does $\sum_{k} 3 \cdot 2^{- k}$ , and by comparison so does $\sum_{k} a_{k}$ .

step 2.2L2L4

4.1

For every $n \in N$ there is an index $k \geq n$ with $s_{k} = - 1$ , namely $k = o_{n}$ , since $o_{n} \geq n$ ; so $q_{k} = 3 / 2$ at some index $k \geq n$ , for every $n$ .

step 3.1L1

5.1

Therefore every tail supremum satisfies $\sup {q_{k} : k \geq n} \geq 3 / 2$ , so $3 / 2$ is a lower bound of the set of tail suprema and ${lim sup}_{k} q_{k} \geq 3 / 2 \geq 1$ .

step 4.1L3

6.1

So $(a_{k})$ has nonzero terms and ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ \geq 1$ , yet $\sum a_{k}$ converges; the claim fails for it and is therefore false.

step 2.1step 5.1step 3.2L6

7.1

Nothing in the ratio test is contradicted: its divergence half requires ${lim inf}_{k} q_{k} > 1$ , and here ${lim inf}_{k} q_{k} \leq 1 / 6 < 1$ , since $q_{k} = 1 / 6$ at indices $e_{n} \geq n$ for every $n$ .

step 6.1step 3.1L1L3L5∎

Remarks

Replacing $lim inf$ by $lim sup$ is already fatal, even with the inequality kept strict. The witness above has ${lim sup}_{k} q_{k} \geq 3 / 2$ , which is strictly greater than $1$ , and its series converges. So the false claim is not rescued by demanding ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ : the two quantities $lim inf$ and $lim sup$ are genuinely different hypotheses here, and only the first one works.
The asymmetry of the ratio test is not an accident of its proof. A large ratio occurring arbitrarily late says only that the terms grow at those steps; it says nothing about their size, because they may have been made very small in between. Only an eventual lower bound on the ratios forces the terms to stay away from $0$ , and that is precisely a hypothesis on $lim inf$ .

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

FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test

Statement

False claim: there is a sequence $(b_{k})$ of reals with $b_{k} > 0$ for every $k \in N$ such that $\sum b_{k}$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series) and such that every sequence $(a_{k})$ of reals with $a_{k} > 0$ for every $k$ and $\sum a_{k}$ divergent satisfies

$b_{k} \leq a_{k} for all k from some index on.$

Such a $(b_{k})$ would be a slowest divergent series of positive terms, and it would give a universal comparison test: a positive series would diverge exactly when its terms eventually dominate those of $(b_{k})$ .

No such sequence exists. The refutation is direct and uses no choice: given any divergent $\sum b_{k}$ with positive terms, the Abel-Dini theorem (For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges) manufactures a divergent series of positive terms whose terms are eventually strictly smaller than the $b_{k}$ , so $(b_{k})$ fails its own defining property.

Facts & Assumptions

Given: An arbitrary sequence $(b_{k})$ of reals with $b_{k} > 0$ for every $k$ and $\sum b_{k}$ divergent; its inclusive partial sums $B_{n} = \sum_{k = 0}^{n} b_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

For a series of nonnegative terms: the partial sums are nondecreasing, and if the series diverges their range is not bounded above and they diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, A nondecreasing sequence that is not bounded above diverges to $+ \infty$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Divergence to $+ \infty$ and to $- \infty$ ).

[L2]

Abel-Dini: if $(a_{k})$ has positive terms and $\sum a_{k}$ diverges, then with $S_{n} = \sum_{k = 0}^{n} a_{k}$ the series $\sum_{n} a_{n} / S_{n}$ diverges (For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges).

[L3]

Order and reciprocals: for $x > 0$ and $y > 1$ one has $0 < x / y < x$ (Inverses of positives are positive, and reciprocation reverses order); and a sum of positive terms is positive (Laws of finite sums and finite products).

[L4]

The refuted claim: some divergent series of positive terms is eventually dominated by every divergent series of positive terms.

Refutation

technique · direct

1.1

Let $(b_{k})$ be any sequence of positive reals with $\sum b_{k}$ divergent, and put $B_{n} = \sum_{k = 0}^{n} b_{k}$ ; every $B_{n}$ is positive, being a sum of positive terms.

givenL3

1.2

Since $\sum b_{k}$ diverges and its terms are nonnegative, its exclusive partial sums $s_{n} = \sum_{k < n} b_{k}$ diverge to $+ \infty$ ; and $B_{n} = s_{n + 1}$ , so $B_{n} \to + \infty$ as well, any index bound for $(s_{n})$ serving for $(B_{n})$ .

givenL1algebra

2.1

Define $c_{n} : = b_{n} / B_{n}$ for $n \in N$ . Each $c_{n}$ is positive, and by Abel-Dini applied to $(b_{k})$ the series $\sum c_{n}$ diverges.

step 1.1L2L3

2.2

Since $B_{n} \to + \infty$ there is $N \in N$ with $B_{n} > 1$ for every $n \geq N$ ; for such $n$ , $c_{n} = b_{n} / B_{n} < b_{n}$ .

step 1.2L3choose

3.1

So $(c_{n})$ is a sequence of positive reals with $\sum c_{n}$ divergent, and there is no index from which $b_{n} \leq c_{n}$ holds onwards: given any $K$ , at every index $n$ that is at least both $K$ and $N$ one has $c_{n} < b_{n}$ .

step 2.1step 2.2

4.1

Therefore the sequence $(b_{k})$ does not have the property demanded of it, and since $(b_{k})$ was an arbitrary divergent series of positive terms, no such sequence exists and the claim is false.

step 3.1L4∎

Remarks

What this rules out. There is no fixed series against which comparison decides divergence for all positive series, so the direct comparison test is unavoidably a family of tests, one for each comparison series, with none of them final. The refutation is constructive in the strong sense: it does not merely show that a slowest series cannot exist, it exhibits, for each candidate, a specific divergent series that beats it.
The scale of tests on this page inherits the same limitation. Ratio, Raabe and Gauss are successive refinements, each deciding series the previous one cannot, and the argument above says the sequence of refinements can never terminate in a universal criterion. What Kummer's test adds is a uniform way of describing the whole family, by naming the weights; it does not escape the obstruction, since each choice of weights is still a comparison against the single series $\sum 1 / ζ_{k}$ .

Series: Convergence and the Nonnegative Tests

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Series, partial sums, convergence and the sum, divergence, and the tail series

Definition

Remarks

A series converges iff each of its tail series converges, and the sum splits as sNsN​ plus the NN-th tail

Statement

Facts & Assumptions

Proof

Remarks

If a series converges then its terms tend to 00

Statement

Facts & Assumptions

Proof

Remarks

A series converges iff for every ε>0ε>0 there is NN with ∣am+1+⋯+an∣<ε∣am+1​+⋯+an​∣<ε for all n>m≥Nn>m≥N

Statement

Facts & Assumptions

Proof

Remarks

Convergent series add and scale termwise

Statement

Facts & Assumptions

Proof

Remarks

If ∑∣ak∣∑∣ak​∣ converges then ∑ak∑ak​ converges

Statement

Facts & Assumptions

Proof

Remarks

For ∣r∣<1∣r∣<1, ∑k≥0rk=1/(1−r)∑k≥0​rk=1/(1−r), and for ∣r∣≥1∣r∣≥1 the series diverges

Statement

Facts & Assumptions

Proof

Remarks

∑(bk−bk+1)∑(bk​−bk+1​) converges iff (bk)(bk​) converges, with sum b0−lim⁡bkb0​−limbk​

Statement

Facts & Assumptions

Proof

Remarks

A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum

Statement

Facts & Assumptions

Proof

Remarks

If 0≤ak≤bk0≤ak​≤bk​ eventually, convergence of ∑bk∑bk​ gives convergence of ∑ak∑ak​, and divergence of ∑ak∑ak​ gives divergence of ∑bk∑bk​

Statement

Facts & Assumptions

Proof

Remarks

For ak,bk>0ak​,bk​>0 with ak/bk→Lak​/bk​→L: if L∈(0,∞)L∈(0,∞) the two series share their behaviour, while L=0L=0 and L=∞L=∞ give one implication each

Statement

Facts & Assumptions

Proof

Remarks

For a nonincreasing nonnegative sequence, ∑ak∑ak​ converges iff ∑2ka2k∑2ka2k​ converges

Statement

Facts & Assumptions

Proof

Remarks

For rational p>0p>0, ∑1/kp∑1/kp converges iff p>1p>1

Statement

Facts & Assumptions

Proof

Remarks

Root test: lim sup⁡∣ak∣1/k<1limsup∣ak​∣1/k<1 gives absolute convergence and hence convergence, >1>1 gives divergence, and =1=1 decides nothing

Statement

Facts & Assumptions

Proof

Remarks

Ratio test: lim sup⁡∣ak+1/ak∣<1limsup∣ak+1​/ak​∣<1 gives absolute convergence and hence convergence, and lim inf⁡∣ak+1/ak∣>1liminf∣ak+1​/ak​∣>1 gives divergence

Statement

Facts & Assumptions

Proof

Remarks

Whenever the ratio test decides, the root test decides the same way, and the converse fails

Statement

A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail

If a series converges then its terms tend to $0$

A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$

If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges

$\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$

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 $\sum b_{k}$

For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each

For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges

For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$

Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing

Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence

Kummer: for positive terms $a_{k}$ and weights $ζ_{k} > 0$ , $lim inf (ζ_{k} a_{k} / a_{k + 1} - ζ_{k + 1}) > 0$ gives convergence, and if $\sum 1 / ζ_{k}$ diverges while that expression is eventually $\leq 0$ the series diverges

Kummer with $ζ_{k} = 1$ recovers the ratio test

Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence

Gauss: for positive terms, if $a_{k} / a_{k + 1} = 1 + h / k + r_{k}$ with $∣ r_{k} ∣ \leq C k^{- 1 - ε}$ for $k \geq 1$ , some constant $C$ and some rational $ε > 0$ , the series converges iff $h > 1$

For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges

FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges

FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges