# Missing Step in Proof that Completeness implies that all Absolutely Convergent Series Converge

I am trying to prove that if $$X$$ is a Banach space, then every absolutely convergent series in $$X$$ converges in $$X$$. My current proof is below but I realize that, in the first paragraph, the first "convergent" is different than the second "convergent". How does this impact my proof, and if it impacts it negatively, how can I remedy it.

We want to prove (Cauchy$$\implies$$Convergent)$$\implies$$(Abs. Conv.$$\implies$$Convergent). This is equivalent to proving Abs. Conv.$$\implies$$Cauchy.

Suppose $$\{x_n\}_\mathbf{N}$$ is a sequence such that the series $$\sum\|x_n\|$$ is absolutely convergent. That is, $$\sum\|x_n\|<\infty$$, i.e., $$\{\sum_{i=1}^n\}_{n\in\mathbf{N}}$$ is convergent. Since, all elements of this sequence are in $$\mathbf{R}$$, this sequence is Cauchy. Then, by definition of Cauchy Sequences,

$$\forall\epsilon>0,\exists N\in\mathbf{N}\ni\forall n,m\ge N:m\ge n,$$

$$\|x_m-x_n\|\le |\|x_m\|-\|x_n\|\le |\|x_m\|+\|x_n\||\le \left|\sum_{i=n}^m\|x_i\|\right|<\epsilon.$$

That is, $$\{x_n\}_{n\in\mathbf{N}}$$ is Cauchy. Therefore, $$X$$ is complete implies that all absolutely convergent series in $$X$$ converges in $$X$$.

• Jun 19, 2020 at 10:40

Since $$\left(\sum_{i=1}^n\|x_i\|\right)_n$$ is a convergent sequence in $$\mathbb{R},$$ it is also Cauchy in $$\mathbb{R}.$$ Let $$\epsilon>0.$$ Then there exists $$n_0 \in \mathbb{N}$$ such that $$\left\vert\sum_{i=1}^m\|x_i\|-\sum_{i=1}^n\|x_i\|\right\vert<\epsilon\;\forall\;m>n\geq n_0$$ and hence $$\left\vert\sum_{i=n+1}^m\|x_i\|\right\vert<\epsilon\;\forall\;m>n\geq n_0.$$
Consider $$\left\Vert\sum_{i=1}^m x_i-\sum_{i=1}^n x_i\right\Vert=\left\Vert \sum_{i=n+1}^m x_i\right\Vert\leq\sum_{i=n+1}^m \|x_i\|<\epsilon\;\forall\;m>n\geq n_0.$$ This implies $$\left(\sum_{i=1}^nx_i\right)_n$$ is Cauchy, hence convergent.