Absolutely Convergent Series is Convergent

Theorem
Let $V$ be a Banach space with norm $\norm {\, \cdot \,}$.

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

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

Proof
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ 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:


 * $\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$

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 $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.

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

Also see

 * Absolutely Convergent Generalized Sum Converges