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

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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]

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 $R$ ). In particular a real $u$ below $\sup {i_{n}}$ is not an upper bound of ${i_{n}}$ ; $i_{n}$ is a lower bound of ${K_{k} : k \geq n}$ ; and $i_{n} \leq K_{n}$ , so $i_{n}$ is not $+ \infty$ .

[L2]

Both limit quantities exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , 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.

Depends on

Used by

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61 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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)

Kummer: for positive terms akak​ and weights ζk>0ζk​>0, lim inf⁡(ζkak/ak+1−ζk+1)>0liminf(ζk​ak​/ak+1​−ζk+1​)>0 gives convergence, and if ∑1/ζk∑1/ζk​ diverges while that expression is eventually ≤0≤0 the series diverges