unconditional convergence

A series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{x}_n$ in a Banach space $X$ is unconditionally convergentif for every permutation $\sigma :\mathbb{N}\to \mathbb{N}$ the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{x}_{\sigma (n)}$converges.

Alternatively, for every chain of finite subsets $S_1\subseteq S_2\subseteq \mathrm{\cdots }$ of $\mathbb{N}$, the partial sums

$$\sum _{k\in S_1}{x}_k,\sum _{k\in S_2}{x}_k,,\mathrm{\dots }$$

converges. The trick to see this equivalence is to realize two facts: 1. every subsequence of a convergent sequence is convergent, and 2. every chain $\{S_i\}$ can be enlarged to a maximal chain $\{T_i\}$, such that $|T_i|=i$. Then the series indexed by $\{S_i\}$ is a subseries indexed by $\{T_i\}$, which is a subseries of a permutation of the original convergent series.

Yet a third equivalent (http://planetmath.org/Equivalent3) definition is given as follows: A series is unconditionally convergent iffor every sequence ${({\epsilon }_n)}_{n=1}^{\mathrm{\infty }}$, with ${\epsilon }_n\in \{\pm 1\}$, theseries ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\epsilon }_n{x}_n$ converges.

Every absolutely convergent series is unconditionally convergent, the converse implication does not hold in general.

When $X={ℝ}^n$ then by a famous theorem of Riemann $(\sum x_n)$ is unconditionally convergent if and only if it is absolutely convergent.