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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 $X$.

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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.

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