Manipulation of Absolutely Convergent Series

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


 * $(1): \quad$ If $\pi: \N \to \N$ is a permutation of $N$, then:


 * $\displaystyle \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty a_{ \pi \left({n}\right) }$


 * $(2): \quad$ If $A \subseteq \N$, and $\chi_A$ is the characteristic function of $A$, then:


 * $\displaystyle \sum_{n \mathop = 1}^\infty a_n \chi_A \left({n}\right) = \sum_{n \mathop \in A} a_n$


 * $(3): \quad$ If $c \in \R$, or $c \in \C$, then:


 * $\displaystyle c \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty c a_n$

Proof of $(1)$
Let $\epsilon > 0$.

From Tail of Convergent Series, it follows that there exists $N \in \N$ such that $\displaystyle \sum_{n \mathop = N}^\infty \left\vert{a_n}\right\vert < \epsilon$.

As a permutation is bijective, 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 $\displaystyle \sum_{n \mathop = 1}^\infty a_n = \lim_{m \to \infty} \sum_{k \mathop = 1}^m a_{\pi \left({k}\right) } = \sum_{k \mathop = 1}^\infty a_{\pi \left({k}\right) }$.

Proof of $(2)$
For all $N \in \N$, we have:


 * $\displaystyle \sum_{n \mathop = 1}^N \left\vert{a_n \chi_A \left({n}\right) }\right\vert \le \sum_{n \mathop = 1}^N \left\vert{a_n}\right\vert \le \sum_{n \mathop = 1}^\infty \left\vert{a_n}\right\vert$

It follows that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{a_n \chi_A \left({n}\right) }\right\vert \le \sum_{n \mathop = 1}^\infty \left\vert{a_n}\right\vert$.

Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n \chi_A \left({n}\right)$ is absolutely convergent.

It follows from $(1)$ that the order of the terms in the series does not matter.

As $\chi_A \left({n}\right) = 0$ for all $n \notin A$, we have:


 * $\displaystyle \sum_{n \mathop = 1}^\infty a_n \chi_A \left({n}\right) = \sum_{n \mathop \in A} a_n$