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

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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.
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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

1.1givenL1

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

1.2givenL4

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 $j \mapsto 1 / ι (j + 1)$ .

2.1step 1.1L1L2choose

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 $∣ a_{k} ∣ = 1 / ι (k + 1) \leq 1 / n < ε$ .

2.2step 1.2L3

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

3.1step 2.1step 2.2L5

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

4.1step 3.1L5∎

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

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

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

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