LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $N \in \mathbb{N}$ , and let $t_j := \sum_{i<j} a_{N+i}$ be the partial sums of the $N$ -th tail series $\sum_{k \ge N} a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

$t_j = s_{j+N} - s_N$ for every $j \in \mathbb{N}$ ;
$\sum a_k$ 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 $m \le n$ 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

1.1

Fix $j \in \mathbb{N}$ and put $n := j + N$ , so that $N \le n$ and $n - N = j$ .

given

1.2

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

givenL2

2.1

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

step 1.1L1

2.2

Suppose $\sum a_k$ converges, say $s_n \to S$ ; then the $N$ -th tail $(s_{j+N})_j$ converges to $S$ .

givenstep 1.2L2

3.1

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

step 2.1algebra

4.1

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 $S - s_N$ .

step 2.2step 3.1L3L4

4.2

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.

step 3.1L3L4

5.1

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.

step 4.2step 1.2L2

6.1

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.

step 4.1step 5.1algebra

7.1

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 $\mathbb{N}$ ; and if some tail series, say the $N$ -th, converges then by claim 2 again $\sum a_k$ converges.

step 6.1∎

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.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 65 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Series (mathematics) (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

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