# Sum of Absolutely Convergent Series

## Theorem

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

Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and:

$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$

## Proof

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

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

$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$

and:

$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$

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

 $\ds \cmod {\sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n - \sum_{n \mathop = 1}^m \paren {a_n + b_n} }$ $=$ $\ds \cmod {\sum_{n \mathop = m + 1}^\infty a_n + \sum_{n \mathop = m + 1}^\infty b_n}$ $\ds$ $\le$ $\ds \sum_{n \mathop = m + 1}^\infty \cmod {a_n} + \sum_{n \mathop = m + 1}^\infty \cmod {b_n}$ by Triangle Inequality $\ds$ $\le$ $\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} + \sum_{n \mathop = M + 1}^\infty \cmod {b_n}$ $\ds$ $<$ $\ds \epsilon$

By definition of convergent series, it follows that:

 $\ds \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$ $=$ $\ds \lim_{m \mathop \to \infty} \sum_{n \mathop = 1}^m \paren {a_n + b_n}$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$

To show that $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, note that:

 $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} + \sum_{n \mathop = 1}^\infty \cmod {b_n}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \paren {\cmod {a_n} + \cmod {b_n} }$ as shown above $\ds$ $\ge$ $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n + b_n}$ by Triangle Inequality

$\blacksquare$