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

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With $a_{k} / b_{k} \to 0$ , convergence of $\sum a_{k}$ does not give convergence of $\sum b_{k}$

Statement refuted

Refuted claim: if $a_{k}, b_{k} > 0$ and $(a_{k} / b_{k})$ converges with $\lim_{k} a_{k} / b_{k} = 0$ , then convergence of $\sum a_{k}$ implies convergence of $\sum b_{k}$ .

Claim 2 of 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 gives the implication in the other direction only: at $L = 0$ , convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ . The claim above reverses it, and the reversal fails. Take

$a_{k} : = \frac{1}{(k + 1)^{2}}, b_{k} : = \frac{1}{k + 1} (k \in N) .$

Both are positive, and $a_{k} / b_{k} = 1 / (k + 1) \to 0$ . But $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ , which converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ at $p = 2$ ), while $\sum b_{k}$ is the harmonic series, which diverges (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping).

The asymmetry is not an artefact of the proof. At $L = 0$ the hypothesis says the $a_{k}$ are eventually much smaller than the $b_{k}$ ; smallness of the $a_{k}$ can never constrain the $b_{k}$ from above, and the witness shows that it does not.

Facts & Assumptions

Given: The sequences $a_{k} : = 1 / ι (k + 1)^{2}$ and $b_{k} : = 1 / ι (k + 1)$ for $k \in N$ , and their quotients $q_{k} = a_{k} / b_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ , Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals $ι (k + 1)$ are positive, so $a_{k}, b_{k} > 0$ ; and reciprocation on the positives is order reversing (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

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 $1 / n < ε$ , Limits and Cauchy sequences of reals).

[L3]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; at $p = 2$ it converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base).

[L4]

The harmonic series $\sum_{k \geq 1} 1 / k$ diverges, and it is the series of $j \mapsto 1 / ι (j + 1)$ (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

Claim 2 of the limit comparison test: with $\lim_{k} a_{k} / b_{k} = 0$ , convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and that is the only implication it supplies in this regime (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).

[L6]

The refuted claim: with $\lim_{k} a_{k} / b_{k} = 0$ , convergence of $\sum a_{k}$ gives convergence of $\sum b_{k}$ .

Counterexample

technique · direct

1.1

Every $a_{k}$ and every $b_{k}$ is positive, so the quotients are defined and the hypotheses of the claim are available for this pair.

givenL1

1.2

The series $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ , the $p$ -series at $p = 2$ , and it converges.

givenL3

1.3

The series $\sum b_{k}$ is the harmonic series, and it diverges.

givenL4

2.1

The quotients are $q_{k} = \frac{1 / ι (k + 1)^{2}}{1 / ι (k + 1)} = \frac{1}{ι (k + 1)}$ , and $(q_{k})$ converges to $0$ : given a rational $ε > 0$ , choose a natural $n \geq 1$ with $1 / n < ε$ , and then $0 < q_{k} \leq 1 / n < ε$ for every $k$ with $k + 1 \geq n$ .

step 1.1L1L2choose

3.1

So $\lim_{k} q_{k} = 0$ and $\sum a_{k}$ converges while $\sum b_{k}$ diverges; the claim is refuted.

step 2.1step 1.2step 1.3L6

4.1

Nothing in the limit comparison test is contradicted: its claim 2 asserts the implication in the opposite direction, and here its hypothesis, convergence of $\sum b_{k}$ , is false.

step 3.1L5∎

Remarks

The same pair also shows the divergence form is one-directional. Read contrapositively, claim 2 says divergence of $\sum a_{k}$ forces divergence of $\sum b_{k}$ . The witness has $\sum b_{k}$ divergent and $\sum a_{k}$ convergent, so divergence of the larger series says nothing about the smaller one, which is the same asymmetry seen from the other side.
The regime $L = + \infty$ fails symmetrically. Exchanging the roles of $(a_{k})$ and $(b_{k})$ in the witness gives $b_{k} / a_{k} \to + \infty$ with $\sum b_{k}$ divergent and $\sum a_{k}$ convergent, so claim 3 of the test is one-directional for the same reason. That reading is immediate from the computation above, the two sequences being the same two.

Depends on

Used by

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Dependency tree · two levels

50 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

Limit comparison test (Wikipedia) (standard reference, not scraped)
APEX Calculus, Section 9.4: Comparison Tests (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

With ak/bk→0ak​/bk​→0, convergence of ∑ak∑ak​ does not give convergence of ∑bk∑bk​