Absolutely Convergent Series is Convergent

Theorem
Let $V$ be a Banach space with norm $\left\Vert{\cdot}\right\Vert$.

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $V$.

Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is convergent.

Proof
That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\displaystyle \sum_{n \mathop = 1}^\infty \left\Vert{a_n}\right\Vert$ converges in $\R$.

Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence is Cauchy Sequence.

Now let $\epsilon > 0$.

Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:


 * $\displaystyle \sum_{k \mathop = n + 1}^m \left\Vert{a_k}\right\Vert = \left\vert{\sum_{k \mathop = 1}^m \left\Vert{a_k}\right\Vert - \sum_{k \mathop = 1}^n \left\Vert{a_k}\right\Vert }\right\vert < \epsilon$

This $N$ exists because the sequence is Cauchy.

Now observe that, for $m \ge n \ge N$, one also has:

It follows that the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.

As $V$ is a Banach space, this implies that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges.

Also see

 * Absolutely Convergent Generalized Sum Converges