LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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An absolutely convergent series has Cauchy partial sums

Statement

Let $V$ be a normed space and let $\sum_{n = 0}^{\infty} x_{n}$ be an absolutely convergent series in $V$ . Then the partial sums form a Cauchy sequence in $V$ .

Facts & Assumptions

Given: A normed space $V$ , a sequence $(x_{n})$ in $V$ , and its partial sums $s_{m} : = \sum_{n < m} x_{n}$ .

[L1]

Absolute convergence means the scalar series $\sum_{n = 0}^{\infty} ∥ x_{n} ∥$ converges (Series and absolute convergence in a normed space).

[L2]

A convergent scalar series has Cauchy partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · direct

1.1L1L2

By [L1] and [L2], for every $ε > 0$ there is $N$ such that $\sum_{ℓ \leq n < m} ∥ x_{n} ∥ < ε$ whenever $m > ℓ \geq N$ .

1.2given

For such $m > ℓ \geq N$ , the tail of the vector partial sums satisfies $s_{m} - s_{ℓ} = \sum_{ℓ \leq n < m} x_{n}$ .

2.1step 1.1step 1.2algebra

Applying the triangle inequality to the finite sum in step 1.2 gives $∥ s_{m} - s_{ℓ} ∥ \leq \sum_{ℓ \leq n < m} ∥ x_{n} ∥ < ε$ .

3.1step 2.1given∎

Since this holds for every $ε > 0$ , the partial sums $(s_{m})$ are Cauchy in the norm metric, which is exactly the claim.

An absolutely convergent series has Cauchy partial sums

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources