TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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

Statement

Let $n \in N$ with $n \geq 1$ and let $(x_{k})$ be a sequence in $R^{n}$ whose series converges absolutely (Series of vectors in $R^{n}$ , 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 $σ : N \to N$ (Injection, surjection, bijection) the rearranged series $\sum x_{σ (k)}$ converges absolutely, with $\sum_{k = 0}^{\infty} x_{σ (k)} = s$ .
Consequently $S (x) = {s}$ : the set of rearrangement sums is a single point.

This is the $R^{n}$ 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 $(R^{n}, d_{2})$ .

Facts & Assumptions

Given: A natural $n \geq 1$ ; a sequence $(x_{k})$ in $R^{n}$ with $\sum ∥ x_{k} ∥_{2}$ convergent; the vector partial sums $s_{N} = \sum_{k < N} x_{k}$ and the real partial sums $T_{N} = \sum_{k < N} ∥ x_{k} ∥_{2}$ ; a bijection $σ$ of $N$ ; a rational $ε > 0$ .

[L1]

Series of vectors, absolute convergence, rearrangement and $S (x)$ (Series of vectors in $R^{n}$ , 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 $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1, Finite sums and finite products, by recursion).

[L2]

The finite triangle inequality for a norm, $N (\sum_{k < p} u_{k}) \leq \sum_{k < p} N (u_{k})$ , and the coordinate bound $∣ y_{j} ∣ \leq ∥ y ∥_{2}$ for $j < n$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ 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 $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ).

[L3]

Splitting of finite sums: for $L \leq N$ , $\sum_{k < N} a_{k} = \sum_{k < L} a_{k} + \sum_{k = L}^{N - 1} a_{k}$ , and the same identity in $R^{n}$ 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 $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1).

[L4]

$(R^{n}, d_{2})$ is complete for $n \geq 1$ , $d_{2} (u, v) = ∥ u - v ∥_{2}$ , and a sequence converging in a metric space is Cauchy (For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm clause 3, Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, $R^{n}$ as the set of functions $n \to R$ , 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 $R$ 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 $R^{n}$ for $n \geq 1$ (For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ 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 $R$ , Limits and Cauchy sequences of reals).

[L6]

The direct comparison test: if $0 \leq a_{k} \leq b_{k}$ from some index on and $\sum b_{k}$ converges, then $\sum a_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ 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 $σ$ of $N$ the series $\sum ∣ a_{σ (k)} ∣$ converges with the same sum as $\sum ∣ a_{k} ∣$ , and $\sum a_{σ (k)}$ 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

1.1L1L3

For $L \leq N$ : $s_{N} - s_{L} = \sum_{k = L}^{N - 1} x_{k}$ and $T_{N} - T_{L} = \sum_{k = L}^{N - 1} ∥ x_{k} ∥_{2}$ , both by splitting, the vector identity being the pointwise reading of the real one.

1.2L4L8

The real sequence $(T_{N})$ converges by hypothesis, hence is Cauchy in $(R, d_{R})$ : for every rational $ε > 0$ there is $K$ with $∣ T_{N} - T_{L} ∣ < ε$ for all $N, L \geq K$ .

1.3L2

For every $j < n$ and every $k$ : $0 \leq ∣ (x_{k})_{j} ∣ \leq ∥ x_{k} ∥_{2}$ .

1.4L7L8L1

Likewise $k \mapsto ∥ x_{σ (k)} ∥_{2}$ is the rearrangement along $σ$ of $k \mapsto ∥ x_{k} ∥_{2}$ , a convergent series of nonnegative terms and therefore absolutely convergent, so $\sum_{k} ∥ x_{σ (k)} ∥_{2}$ converges; that is, $\sum x_{σ (k)}$ converges absolutely.

2.1step 1.1L2

Hence $∥ s_{N} - s_{L} ∥_{2} \leq \sum_{k = L}^{N - 1} ∥ x_{k} ∥_{2} = T_{N} - T_{L}$ by the finite triangle inequality.

2.2step 1.3L6L8

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.

3.1step 2.1step 1.2L4

By steps 2.1 and 1.2, for $N \geq L \geq K$ we get $d_{2} (s_{N}, s_{L}) = ∥ s_{N} - s_{L} ∥_{2} \leq ∣ T_{N} - T_{L} ∣ < ε$ , and the same bound with $N$ and $L$ exchanged; so $(s_{N})$ is Cauchy in $(R^{n}, d_{2})$ .

3.2step 2.2L7

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

4.1step 3.1L4

Since $(R^{n}, d_{2})$ is complete, the Cauchy sequence $(s_{N})$ converges; that is, $\sum x_{k}$ converges, which is clause 1. Write $s$ for its sum.

5.1step 4.1step 3.2step 1.4L5

By clause 1 applied to the sequence $k \mapsto x_{σ (k)}$ , which converges absolutely by step 1.4, the series $\sum x_{σ (k)}$ 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.

6.1step 4.1step 5.1L1∎

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

Remarks

Two independent routes to clause 1 are available and only one is used. The proof above uses completeness of $(R^{n}, d_{2})$ together with the finite triangle inequality. The alternative is to run step 2.2 first and reassemble by For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ 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 $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ ) 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 $S (x)$ looks like when the series converges without converging absolutely is taken up in The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$ , which proves a containment and no more; see Conventions of this page, the standing $n \geq 1$ hypothesis, and what is taken up elsewhere in the reading order for exactly what this page does and does not settle.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

An absolutely convergent series in RnRn converges, and every rearrangement converges to the same sum

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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