TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Kummer: for positive terms $a_k$ and weights $\zeta_k > 0$ , $\liminf(\zeta_k a_k/a_{k+1} - \zeta_{k+1}) > 0$ gives convergence, and if $\sum 1/\zeta_k$ diverges while that expression is eventually $\le 0$ the series diverges

Statement

Let $(a_k)$ and $(\zeta_k)$ be sequences of reals with

$a_k > 0 \quad \text{and} \quad \zeta_k > 0 \qquad \text{for every } k \in \mathbb{N},$

and define Kummer's expression

$K_k \;:=\; \zeta_k \frac{a_k}{a_{k+1}} \;-\; \zeta_{k+1} \qquad (k \in \mathbb{N}),$

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

if $\liminf_{k} K_k > 0$ then $\sum a_k$ converges;
if $\sum 1/\zeta_k$ diverges and $K_k \le 0$ 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 $\zeta_k = 1$ .

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

Facts & Assumptions

Given: Sequences $(a_k)$ , $(\zeta_k)$ of reals with $a_k > 0$ and $\zeta_k > 0$ for every $k$ ; Kummer's expression $K_k = \zeta_k a_k/a_{k+1} - \zeta_{k+1}$ ; the auxiliary sequence $b_k := \zeta_k a_k$ , which is positive; and the tail infima $i_n = \inf\{K_k : k \ge n\}$ taken in $\overline{\mathbb{R}}$ , so that $\liminf_k K_k = \sup\{i_n : n \in \mathbb{N}\}$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L1]

Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound there (Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ ). In particular a real $u$ below $\sup\{i_n\}$ is not an upper bound of $\{i_n\}$ ; $i_n$ is a lower bound of $\{K_k : k \ge n\}$ ; and $i_n \le K_n$ , so $i_n$ is not $+\infty$ .

[L2]

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

[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 $0 \le x_k \le y_k$ from some index on and $\sum y_k$ converges then $\sum x_k$ converges (If $0 \le a_k \le b_k$ 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 $c \ne 0$ , $\sum c\,x_k$ 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 $x \le y$ implying $1/y \le 1/x$ for positive $x, y$ (Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct

1.1

Suppose $\liminf_k K_k > 0$ . The real $0$ is then not an upper bound of $\{i_n\}$ , so there is $N \in \mathbb{N}$ with $i_N > 0$ .

givenL1L2choose

1.2

Suppose now that $\sum 1/\zeta_k$ diverges and that there is $N$ with $K_k \le 0$ for every $k \ge N$ .

given

2.1

Since $i_N \le K_N$ and $i_N > 0$ , the value $i_N$ is a real number; put $c := i_N > 0$ , so that $K_k \ge c$ for every $k \ge N$ .

step 1.1L1

2.2

Multiplying $K_k \le 0$ by $a_{k+1} > 0$ gives $\zeta_k a_k \le \zeta_{k+1} a_{k+1}$ , that is $b_k \le b_{k+1}$ , for every $k \ge N$ .

step 1.2givenalgebra

3.1

Multiplying $K_k \ge c$ by $a_{k+1} > 0$ gives $\zeta_k a_k - \zeta_{k+1} a_{k+1} \ge c\,a_{k+1}$ , that is $b_k - b_{k+1} \ge c\,a_{k+1} > 0$ , for every $k \ge N$ .

step 2.1givenalgebra

3.2

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

step 2.2L7

4.1

Hence $b_{N+j} \ge b_{N+j+1}$ for every $j \in \mathbb{N}$ , 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 $\zeta_m a_m \ge b_N > 0$ for every $m \ge N$ , and dividing by $\zeta_m > 0$ gives $a_m \ge b_N \cdot (1/\zeta_m) > 0$ .

step 3.2givenL7

5.1

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

step 4.1L3L4

5.2

Since $\sum 1/\zeta_k$ diverges and $b_N \ne 0$ , the series $\sum b_N (1/\zeta_k)$ diverges.

step 1.2step 4.2L6

6.1

By step 3.1 we have $0 \le c\,a_{N+j+1} \le b_{N+j} - b_{N+j+1}$ for every $j$ , so $\sum_{j} c\,a_{N+j+1}$ converges by comparison, and since $c \ne 0$ 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 $0 \le b_N(1/\zeta_m) \le a_m$ for $m \ge N$ , comparison would make $\sum b_N(1/\zeta_k)$ 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 $\zeta_k = 1$ 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 $K_k = -2 - 1 = -3 \le 0$ at every index; and $\sum 1/\zeta_k = \sum 1$ 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 $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $\zeta_k = 1$ 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 $(\zeta_k)$ , and the strength of the resulting test is exactly the strength of the divergent comparison series $\sum 1/\zeta_k$ 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 $\sum 1/\zeta_k$ 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.

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Direct dependencies and their dependencies through the next three levels: 109 results over 28 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Convergence tests (Wikipedia) (standard reference, not scraped)
K. Knopp, Theory and Application of Infinite Series, Ch. IX (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)
Binghamton University notes on Kummer, Raabe, and Gauss tests (standard reference, not scraped)