Manipulation of Absolutely Convergent Series/Characteristic Function

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Theorem

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


Let $A \subseteq \N$.

Then:

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

where $\chi_A$ is the characteristic function of $A$.


Proof

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 Manipulation of Absolutely Convergent Series: Permutation 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:

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

$\blacksquare$