# Manipulation of Absolutely Convergent Series/Permutation

## Theorem

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a real or complex series that is absolutely convergent.

If $\pi: \N \to \N$ is a permutation of $N$, then:

$\displaystyle \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty a_{\pi \left({n}\right)}$

## Proof

Let $\epsilon > 0$.

From Tail of Convergent Series tends to Zero, it follows that there exists $N \in \N$ such that:

$\displaystyle \sum_{n \mathop = N}^\infty \left\vert{a_n}\right\vert < \epsilon$

By definition, a permutation is bijective.

Hence we can find $M \in \N$ such that:

$\left\{ {1, \ldots, N-1}\right\} \subseteq \left\{ {\pi \left({1}\right), \ldots, \pi \left({M}\right) }\right\}$

Let $m \in \N$, and put $B = \left\{ {n \in N: \pi^{-1} \left({n}\right) > m}\right\}$.

For all $m \ge M$, it follows that:

 $\displaystyle \left\vert{ \sum_{n \mathop = 1}^\infty a_n - \sum_{k \mathop = 1}^m a_{\pi \left({k}\right) } }\right\vert$ $=$ $\displaystyle \left\vert{ \sum_{n \mathop = 1}^\infty a_n \chi_B \left({n}\right) }\right\vert$ where $\chi_B$ is the characteristic function of $B$ $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{a_n}\right\vert \chi_B \left({n}\right)$ by Triangle Inequality $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = N}^\infty \left\vert{a_n}\right\vert$ as $\chi_B = 0$ for all $n < N$ $\displaystyle$ $<$ $\displaystyle \epsilon$

By definition of convergent series, it follows that:

$\displaystyle \sum_{n \mathop = 1}^\infty a_n = \lim_{m \to \infty} \sum_{k \mathop = 1}^m a_{\pi \left({k}\right)} = \sum_{k \mathop = 1}^\infty a_{\pi \left({k}\right)}$

$\blacksquare$