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:

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