Absolutely convergent series in a Banach space converge

Statement

Let $(V, \|\cdot\|)$ be a Banach space and let $(x_n)_{n \ge 1}$ be a sequence in $V$ with $\sum_{n \ge 1} \|x_n\| < \infty$. Then the partial sums $s_N = \sum_{n=1}^{N} x_n$ converge to a point $s \in V$, and $\|s\| \le \sum_{n \ge 1} \|x_n\|$.

Remarks

Proof: for $M > N$, the triangle inequality gives $\|s_M - s_N\| \le \sum_{n=N+1}^{M} \|x_n\|$. This tends to $0$, so $(s_N)$ is a Cauchy sequence, and it converges because $V$ is complete. The converse holds: a normed space in which each absolutely convergent series converges is a Banach space. For $V = \mathbb{R}$, this is the fact that absolutely convergent real series converge.

Explanations

L1 human

If the lengths of the terms have a finite sum, then the series converges in a Banach space. This is a practical test for completeness, and it is often used to build solutions as sums of series.