Manipulation of Absolutely Convergent Series/Permutation

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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:

$\displaystyle \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty a_{\map \pi n}$


Proof

Let $\epsilon > 0$.

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

$\displaystyle \sum_{n \mathop = N}^\infty \size {a_n} < \epsilon$


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:

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

By definition of convergent series, it follows that:

$\displaystyle \sum_{n \mathop = 1}^\infty a_n = \lim_{m \mathop \to \infty} \sum_{k \mathop = 1}^m a_{\map \pi k} = \sum_{k \mathop = 1}^\infty a_{\map \pi k}$

$\blacksquare$


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