Manipulation of Absolutely Convergent Series/Permutation

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

If $\pi: \N \to \N$ is a permutation of $N$, then:
 * $\ds \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty a_{\map \pi n}$

Proof
Let $\epsilon > 0$.

From Tail of Convergent Series tends to Zero, it follows that there exists $N \in \N$ such that:
 * $\ds \sum_{n \mathop = N}^\infty \size {a_n} < \epsilon$

By definition, a permutation is bijective.

Hence we can find $M \in \N$ such that:
 * $\set {1, \ldots, N - 1} \subseteq \set {\map \pi 1, \ldots, \map \pi M}$

Let $m \in \N$, and put $B = \set {n \in N: \map {\pi^{-1} } n > m}$.

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

By definition of convergent series, it follows that:
 * $\ds \sum_{n \mathop = 1}^\infty a_n = \lim_{m \mathop \to \infty} \sum_{k \mathop = 1}^m a_{\map \pi k} = \sum_{k \mathop = 1}^\infty a_{\map \pi k}$