Sum of Absolutely Convergent Series

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

Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \left({a_n + b_n}\right)$ is absolutely convergent, and:


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

Proof
Let $\epsilon \in \R_{>0}$.

From Tail of Convergent Series, it follows that there exists $M \in \N$ such that $\displaystyle \sum_{n \mathop = M+1}^\infty \left\vert{a_n}\right\vert < \dfrac \epsilon 2$, and $\displaystyle \sum_{n \mathop = M+1}^\infty \left\vert{b_n}\right\vert < \dfrac \epsilon 2$.

For all $m \ge M$, it follows that:

By definition of convergent series, it follows that:


 * $\displaystyle \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n = \lim_{m \to \infty} \sum_{n \mathop = 1}^m \left({a_n + b_n}\right) = \sum_{n \mathop = 1}^\infty \left({a_n + b_n}\right)$.

To show that $\displaystyle \sum_{n \mathop = 1}^\infty \left({a_n + b_n}\right)$ is absolutely convergent, note that: