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

How statement and proof provenance work

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.

If a series converges then its terms tend to $0$

Statement

Let $(a_{k})$ be a sequence of reals. If the series $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the sequence $(a_{k})$ converges (Limits and Cauchy sequences of reals), with

$\lim_{k} a_{k} = 0.$

Equivalently, in the contrapositive form in which the lemma is almost always used: if $(a_{k})$ does not converge to $0$ then $\sum a_{k}$ diverges.

The same statement holds for a series $\sum_{k \geq m} a_{k}$ with a general starting index, with the conclusion that $(a_{j + m})_{j \in N}$ converges to $0$ : that series is by definition the series of the sequence $b_{j} = a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), so it is the statement above applied to $(b_{j})$ .

The converse is false; it is refuted by FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals whose series converges, with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[A1]

$(s_{n})$ converges to some real $S$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L1]

The recursion clause of the finite sum: $s_{n + 1} = s_{n} + a_{n}$ for every $n \in N$ (Finite sums and finite products, by recursion, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Convergence depends only on the tail: if $(x_{n})$ converges to $x$ then so does its $1$ -st tail $(x_{n + 1})_{n}$ (Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

Algebra of limits: if $x_{n} \to x$ and $y_{n} \to y$ then $x_{n} - y_{n} \to x - y$ (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct

1.1A1L2

The sequence $(s_{n + 1})_{n \in N}$ is the $1$ -st tail of $(s_{n})$ , and it converges to $S$ .

1.2L1algebra

From the recursion clause, $a_{n} = s_{n + 1} - s_{n}$ for every $n \in N$ .

2.1step 1.1step 1.2A1L3

Hence $(a_{n})$ is the difference of two sequences converging to $S$ , so it converges to $S - S = 0$ .

3.1step 2.1∎

That is the claim, and its contrapositive is the assertion that a series whose terms do not tend to $0$ diverges.

Remarks

This is a necessary condition and never a sufficient one. It rules a series out; it never rules one in. The harmonic series has terms tending to $0$ and diverges, which is exactly the content of FALSE: if $a_{k} \to 0$ then $\sum a_{k}$ converges.
What the proof actually uses. Only that the partial sums converge and that consecutive partial sums differ by a term. No sign hypothesis is placed on $(a_{k})$ , and none is available at this point on the page.

If a series converges then its terms tend to $0$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If a series converges then its terms tend to 00

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If a series converges then its terms tend to $0$