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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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]

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).

[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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

49 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

Term test (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

FALSE: if ak→0ak​→0 then ∑ak∑ak​ converges