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

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A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail

Statement

Let $(a_{k})$ be a sequence of reals with partial sums $s_{n} = \sum_{k < n} a_{k}$ , let $N \in N$ , and let $t_{j} : = \sum_{i < j} a_{N + i}$ be the partial sums of the $N$ -th tail series $\sum_{k \geq N} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

$t_{j} = s_{j + N} - s_{N}$ for every $j \in N$ ;
$\sum a_{k}$ converges if and only if its $N$ -th tail series converges, and in that case $\sum_{k = 0}^{\infty} a_{k} = s_{N} + \sum_{k = N}^{\infty} a_{k};$
hence the following are equivalent: $\sum a_{k}$ converges; every tail series of $\sum a_{k}$ converges; some tail series of $\sum a_{k}$ converges.

In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals, a natural number $N$ , the partial sums $s_{n} = \sum_{k < n} a_{k}$ and the partial sums $t_{j} = \sum_{i < j} a_{N + i}$ of the $N$ -th tail series (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Splitting of finite sums: if $m \leq n$ then $\sum_{k < n} a_{k} = \sum_{k < m} a_{k} + \sum_{k = m}^{n - 1} a_{k}$ (Laws of finite sums and finite products), and $\sum_{k = m}^{n - 1} a_{k}$ is by definition $\sum_{i < n - m} a_{m + i}$ (Finite sums and finite products, by recursion).

[L2]

Convergence depends only on the tail: a sequence $(x_{n})$ converges to $x$ if and only if its $N$ -th tail $(x_{j + N})_{j}$ converges to $x$ (Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

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

[L4]

A constant sequence $y_{j} = c$ converges to $c$ , immediately from the definition of a limit (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1given

Fix $j \in N$ and put $n : = j + N$ , so that $N \leq n$ and $n - N = j$ .

1.2givenL2

The family $(s_{j + N})_{j \in N}$ is the $N$ -th tail of the sequence $(s_{n})$ of partial sums.

2.1step 1.1L1

Splitting the partial sum $s_{n}$ at $N$ gives $s_{n} = \sum_{k < N} a_{k} + \sum_{k = N}^{n - 1} a_{k} = s_{N} + \sum_{i < j} a_{N + i} = s_{N} + t_{j}$ .

2.2givenstep 1.2L2

Suppose $\sum a_{k}$ converges, say $s_{n} \to S$ ; then the $N$ -th tail $(s_{j + N})_{j}$ converges to $S$ .

3.1step 2.1algebra

Claim 1 follows: $t_{j} = s_{j + N} - s_{N}$ for every $j \in N$ .

4.1step 2.2step 3.1L3L4

The constant sequence with value $s_{N}$ converges to $s_{N}$ , so $t_{j} = s_{j + N} - s_{N} \to S - s_{N}$ , and the $N$ -th tail series converges with sum $S - s_{N}$ .

4.2step 3.1L3L4

Conversely, suppose the $N$ -th tail series converges, say $t_{j} \to T$ ; then $s_{j + N} = s_{N} + t_{j} \to s_{N} + T$ by the same two rules.

5.1step 4.2step 1.2L2

So the $N$ -th tail of $(s_{n})$ converges to $s_{N} + T$ , hence $(s_{n})$ itself converges to $s_{N} + T$ and $\sum a_{k}$ converges.

6.1step 4.1step 5.1algebra

Claim 2 follows: the two convergences are equivalent, and when they hold the sums are related by $S = s_{N} + T$ , which is the displayed identity.

7.1step 6.1∎

Claim 3 follows as well: if $\sum a_{k}$ converges then by claim 2 every tail series converges; every tail series converging trivially gives some tail series converging, the family of tails being indexed by all of $N$ ; and if some tail series, say the $N$ -th, converges then by claim 2 again $\sum a_{k}$ converges.

Remarks

Where the hypothesis-free character comes from. Nothing here assumes the terms have a sign or that any series converges: claim 1 is an identity between finite sums, valid always, and claims 2 and 3 are read off from it by two standard limit rules. That is why the lemma may be used to move between a series and its tails inside proofs that have not yet decided the convergence question.
The sum does change. Only the fact of convergence is tail invariant. The identity in claim 2 is the exact bookkeeping: discarding the first $N$ terms lowers the sum by $s_{N}$ , and no more.

A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A series converges iff each of its tail series converges, and the sum splits as sNsN​ plus the NN-th tail

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail