Manipulation of Absolutely Convergent Series/Permutation
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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:
\(\ds \size {\sum_{n \mathop = 1}^\infty a_n - \sum_{k \mathop = 1}^m a_{\map \pi k} }\) | \(=\) | \(\ds \size {\sum_{n \mathop = 1}^\infty a_n \map {\chi_B} n}\) | where $\chi_B$ is the characteristic function of $B$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 1}^\infty \size {a_n} \map {\chi_B} n\) | 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:
- $\ds \sum_{n \mathop = 1}^\infty a_n = \lim_{m \mathop \to \infty} \sum_{k \mathop = 1}^m a_{\map \pi k} = \sum_{k \mathop = 1}^\infty a_{\map \pi k}$
$\blacksquare$
Sources
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.3$: Operations with series