Manipulation of Absolutely Convergent Series/Permutation

From ProofWiki
Jump to navigation Jump to search

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:

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

$\blacksquare$


Sources