TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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An absolutely convergent series in $\mathbb{R}^n$ converges, and every rearrangement converges to the same sum

Statement

Let $n \in \mathbb{N}$ with $n \ge 1$ and let $(x_k)$ be a sequence in $\mathbb{R}^{n}$ whose series converges absolutely (Series of vectors in $\mathbb{R}^n$ , absolute convergence, rearrangement, and the set of rearrangement sums). Then:

$\sum x_k$ converges; write $s := \sum_{k=0}^{\infty}x_k$ .
For every bijection $\sigma : \mathbb{N} \to \mathbb{N}$ (Injection, surjection, bijection) the rearranged series $\sum x_{\sigma(k)}$ converges absolutely, with $\sum_{k=0}^{\infty}x_{\sigma(k)} = s$ .
Consequently $\mathcal{S}(x) = \{s\}$ : the set of rearrangement sums is a single point.

This is the $\mathbb{R}^{n}$ analogue of the published one-dimensional statements, not a generalisation of their proofs. If $\sum |a_k|$ converges then $\sum a_k$ converges and Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum are proved on the real line; everything below reduces to them coordinatewise, or to completeness of $(\mathbb{R}^{n},d_2)$ .

Facts & Assumptions

Given: A natural $n \ge 1$ ; a sequence $(x_k)$ in $\mathbb{R}^{n}$ with $\sum \lVert x_k\rVert_2$ convergent; the vector partial sums $s_N = \sum_{k<N}x_k$ and the real partial sums $T_N = \sum_{k<N}\lVert x_k\rVert_2$ ; a bijection $\sigma$ of $\mathbb{N}$ ; a rational $\varepsilon > 0$ .

[L1]

Series of vectors, absolute convergence, rearrangement and $\mathcal{S}(x)$ (Series of vectors in $\mathbb{R}^n$ , absolute convergence, rearrangement, and the set of rearrangement sums); partial sums are computed pointwise, $(s_N)(j) = \sum_{k<N}(x_k)_j$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ clause 1, Finite sums and finite products, by recursion).

[L2]

The finite triangle inequality for a norm, $N(\sum_{k<p}u_k) \le \sum_{k<p}N(u_k)$ , and the coordinate bound $|y_j| \le \lVert y\rVert_2$ for $j<n$ (The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clauses 1 and 3, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ ).

[L3]

Splitting of finite sums: for $L \le N$ , $\sum_{k<N}a_k = \sum_{k<L}a_k + \sum_{k=L}^{N-1}a_k$ , and the same identity in $\mathbb{R}^{n}$ read pointwise (Laws of finite sums and finite products clause 3, Finite sums and finite products, by recursion, The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ clause 1).

[L4]

$(\mathbb{R}^{n},d_2)$ is complete for $n \ge 1$ , $d_2(u,v) = \lVert u-v\rVert_2$ , and a sequence converging in a metric space is Cauchy (For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 3, Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$ , and the induced metrics are exactly $d_1$ , $d_2$ and $d_\infty$ of the published metric-spaces page, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, Complete metric space: every Cauchy sequence converges in the space, Cauchy sequence in a metric space, Every convergent sequence in a metric space is Cauchy, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded).

[L5]

Componentwise convergence in $\mathbb{R}^{n}$ for $n \ge 1$ (For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 1, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Limits and Cauchy sequences of reals).

[L6]

The direct comparison test: if $0 \le a_k \le b_k$ from some index on and $\sum b_k$ converges, then $\sum a_k$ converges (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

Dirichlet's rearrangement theorem: if $\sum a_k$ converges absolutely then for every bijection $\sigma$ of $\mathbb{N}$ the series $\sum |a_{\sigma(k)}|$ converges with the same sum as $\sum|a_k|$ , and $\sum a_{\sigma(k)}$ converges with the same sum as $\sum a_k$ (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Absolutely convergent and conditionally convergent series, and the general starting index).

[L8]

Absolute convergence implies convergence for real series, and a convergent series of nonnegative terms is absolutely convergent, its terms being their own absolute values (If $\sum |a_k|$ converges then $\sum a_k$ converges, Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct

1.1

For $L \le N$ : $s_N - s_L = \sum_{k=L}^{N-1}x_k$ and $T_N - T_L = \sum_{k=L}^{N-1}\lVert x_k\rVert_2$ , both by splitting, the vector identity being the pointwise reading of the real one.

L1L3

1.2

The real sequence $(T_N)$ converges by hypothesis, hence is Cauchy in $(\mathbb{R},d_{\mathbb{R}})$ : for every rational $\varepsilon>0$ there is $K$ with $|T_N - T_L| < \varepsilon$ for all $N,L \ge K$ .

L4L8

1.3

For every $j<n$ and every $k$ : $0 \le |(x_k)_j| \le \lVert x_k\rVert_2$ .

1.4

Likewise $k \mapsto \lVert x_{\sigma(k)}\rVert_2$ is the rearrangement along $\sigma$ of $k \mapsto \lVert x_k\rVert_2$ , a convergent series of nonnegative terms and therefore absolutely convergent, so $\sum_k\lVert x_{\sigma(k)}\rVert_2$ converges; that is, $\sum x_{\sigma(k)}$ converges absolutely.

L7L8L1

2.1

Hence $\lVert s_N - s_L\rVert_2 \le \sum_{k=L}^{N-1}\lVert x_k\rVert_2 = T_N - T_L$ by the finite triangle inequality.

step 1.1L2

2.2

By step 1.3 and the comparison test, the real series $\sum_k |(x_k)_j|$ converges for every $j<n$ ; so each coordinate series $\sum_k (x_k)_j$ converges absolutely.

step 1.3L6L8

3.1

By steps 2.1 and 1.2, for $N \ge L \ge K$ we get $d_2(s_N,s_L) = \lVert s_N-s_L\rVert_2 \le |T_N-T_L| < \varepsilon$ , and the same bound with $N$ and $L$ exchanged; so $(s_N)$ is Cauchy in $(\mathbb{R}^{n},d_2)$ .

step 2.1step 1.2L4

3.2

Fix a bijection $\sigma$ . For every $j<n$ the sequence $k \mapsto (x_{\sigma(k)})_j$ is the rearrangement along $\sigma$ of the sequence $k \mapsto (x_k)_j$ ; by step 2.2 the latter series converges absolutely, so Dirichlet's theorem gives that $\sum_k (x_{\sigma(k)})_j$ converges with the same sum as $\sum_k (x_k)_j$ .

step 2.2L7

4.1

Since $(\mathbb{R}^{n},d_2)$ is complete, the Cauchy sequence $(s_N)$ converges; that is, $\sum x_k$ converges, which is clause 1. Write $s$ for its sum.

step 3.1L4

5.1

By clause 1 applied to the sequence $k \mapsto x_{\sigma(k)}$ , which converges absolutely by step 1.4, the series $\sum x_{\sigma(k)}$ converges; and by step 3.2 each coordinate of its sum equals the corresponding coordinate of $s$ , so its sum is $s$ . This is clause 2.

step 4.1step 3.2step 1.4L5

6.1

By clause 2 every rearrangement of $\sum x_k$ converges to $s$ , and the identity bijection shows $s \in \mathcal{S}(x)$ ; so $\mathcal{S}(x) = \{s\}$ , which is clause 3.

step 4.1step 5.1L1∎

Remarks

Two independent routes to clause 1 are available and only one is used. The proof above uses completeness of $(\mathbb{R}^{n},d_2)$ together with the finite triangle inequality. The alternative is to run step 2.2 first and reassemble by For $n \ge 1$ a sequence in $\mathbb{R}^n$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $\mathbb{R}^n$ is complete in every norm clause 1, using If $\sum |a_k|$ converges then $\sum a_k$ converges on each coordinate. The two give the same theorem; mixing them would prove clause 1 twice.
The published Cauchy criterion for series (A series converges iff for every $\varepsilon > 0$ there is $N$ with $|a_{m+1} + \dots + a_n| < \varepsilon$ for all $n > m \ge N$ ) is the standard packaging of step 1.2 and would serve in its place; the proof uses the plainer statement that a convergent real sequence is Cauchy, so that the index bookkeeping stays visible.
Clause 3 is the half of the rearrangement question this theorem settles. For an absolutely convergent series the set of rearrangement sums is as small as it can be. What $\mathcal{S}(x)$ looks like when the series converges without converging absolutely is taken up in The set of rearrangement sums of a convergent series in $\mathbb{R}^n$ is a nonempty subset of the affine subspace $s + \Gamma^{\perp}$ , which proves a containment and no more; see Conventions of this page, the standing $n \ge 1$ hypothesis, and what is taken up elsewhere in the reading order for exactly what this page does and does not settle.

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 206 results over 38 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

Absolute convergence (Wikipedia) (standard reference, not scraped)
Riemann series theorem (Wikipedia) (standard reference, not scraped)
T. Banakh, A Simple Inductive Proof of the Levy-Steinitz Theorem (standard reference, not scraped)

An absolutely convergent series in Rn\mathbb{R}^nRn converges, and every rearrangement converges to the same sum