Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Necessary Condition

Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

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

Suppose $X$ is a Banach space.

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

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 $\R$.

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:

$\ds \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}$

$=$

$\ds \norm {\sum_{k \mathop = n + 1}^m a_k}$

$\ds $

$\le$

$\ds \sum_{k \mathop = n + 1}^m \norm {a_k}$

Triangle inequality for $\norm {\, \cdot \,}$

$\ds $

$<$

$\ds \epsilon$

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

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

$\blacksquare$

2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces