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

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Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum

Statement

Let $(a_{k})$ be a sequence of reals whose series converges (Series, partial sums, convergence and the sum, divergence, and the tail series), with sum $S$ , and let $n : N \to N$ be strictly increasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) with $n_{0} = 0$ . Define the blocks

$B_{j} : = \sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k} (j \in N),$

each a finite sum of $n_{j + 1} - n_{j} \geq 1$ consecutive terms (Finite sums and finite products, by recursion). Then $\sum B_{j}$ converges, with

$\sum_{j = 0}^{\infty} B_{j} = S .$

The proof shows more, and the extra is what makes the theorem trivial once seen: the $m$ -th partial sum of $\sum B_{j}$ is exactly $s_{n_{m}}$ , the $n_{m}$ -th partial sum of $\sum a_{k}$ . Grouping does not produce a new series so much as a subsequence of the old partial sums.

The converse fails. A grouped series may converge while the original diverges, and FALSE: if some grouping of a series converges then the series itself converges records that, with the witness on the companion page. What the theorem needs is convergence of $\sum a_{k}$ as a hypothesis, and $n_{0} = 0$ , without which the first block would omit the terms before $n_{0}$ .

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $\sum a_{k}$ convergent of sum $S$ and partial sums $s_{n} = \sum_{k < n} a_{k}$ ; a strictly increasing $n : N \to N$ with $n_{0} = 0$ ; and the blocks $B_{j} = \sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ .

[L1]

Finite sums: $\sum_{k < 0} x_{k} = 0$ and $\sum_{k < m + 1} x_{k} = \sum_{k < m} x_{k} + x_{m}$ (Finite sums and finite products, by recursion).

[L2]

Splitting: 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).

[L3]

The partial sums of a series and the meaning of its sum (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

A subsequence of a convergent sequence converges to the same limit; a subsequence is indexed by a strictly increasing map $N \to N$ (Subsequences inherit the limit, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L5]

The principle of induction on $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

Since $n$ is strictly increasing, $n_{j} < n_{j + 1}$ for every $j$ , so each block $B_{j}$ is a finite sum over a nonempty range of indices and is a well-determined real.

givenL1L2

1.2

The map $m \mapsto n_{m}$ is strictly increasing, so $(s_{n_{m}})_{m}$ is a subsequence of the convergent sequence $(s_{n})$ and therefore converges to $S$ .

givenL3L4

2.1

An induction on $m$ gives $\sum_{j < m} B_{j} = s_{n_{m}}$ for every $m \in N$ : at $m = 0$ the left side is the empty sum $0$ and the right side is $s_{n_{0}} = s_{0} = 0$ ; and if $\sum_{j < m} B_{j} = s_{n_{m}}$ then $\sum_{j < m + 1} B_{j} = s_{n_{m}} + B_{m} = \sum_{k < n_{m}} a_{k} + \sum_{k = n_{m}}^{n_{m + 1} - 1} a_{k} = \sum_{k < n_{m + 1}} a_{k} = s_{n_{m + 1}}$ , the middle equality being splitting at $n_{m} \leq n_{m + 1}$ .

givenstep 1.1L1L2L5

3.1

By step 2.1 the partial sums of $\sum B_{j}$ are precisely the terms $s_{n_{m}}$ , so $\sum B_{j}$ converges with sum $S$ .

step 2.1step 1.2L3∎

Remarks

Why $n_{0} = 0$ is a hypothesis and not a normalisation. If $n_{0} > 0$ the same computation gives $\sum_{j < m} B_{j} = s_{n_{m}} - s_{n_{0}}$ , so the grouped series converges to $S - s_{n_{0}}$ : the terms $a_{0}, \dots, a_{n_{0} - 1}$ are simply omitted. The theorem as stated is the case where nothing is omitted.
Blocks may be as long as one likes, and the theorem is indifferent. No bound on $n_{j + 1} - n_{j}$ is assumed, and none is needed: the argument never looks inside a block. This is exactly what fails in the converse direction, where the cancellation hidden inside long blocks is what the grouped series cannot see.
The result also gives the associativity one expects of a convergent series. Any two groupings of a convergent series have the same sum, both being $S$ ; so one may insert brackets at will, though never remove them (FALSE: if some grouping of a series converges then the series itself converges).

Depends on

Used by

(1-1) + (1-1) + … converges to 0 while ∑ₖ (-1)ᵏ diverges Counterexample
FALSE: if some grouping of a series converges then the series itself converges False statement
Euler's formula: exp(iθ)=cosθ+i sinθ for every real θ Theorem

Dependency tree · two levels

30 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

Series (mathematics) (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §7.4 (standard reference, not scraped)
W. Fisher, Introduction to Analysis (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis, Chapter 4 (standard reference, not scraped)

Grouping: if ∑ak∑ak​ converges and (nj)(nj​) is strictly increasing with n0=0n0​=0, the series of blocks ∑k=njnj+1−1ak∑k=nj​nj+1​−1​ak​ converges to the same sum