TITLE: Proving that $X$ is a Banach space iff convergence of $\sum\|x_n\|$ implies convergence of $\sum x_n$
QUESTION [8 upvotes]: The following is an Exercise of Conway's Functional Analysis. 
Prove that $X$ is a $\,Banach$ space iff whenever $\{x_n\}$ is a sequence in $X$, such that $\sum \| x_n \| < \infty$, then $\sum x_n$ converges. 
I easily can show that if $X$ is a Banach space then $\sum x_n$ converges. My problem is showing that $X$ is a Banach space. For this I suppose that $\{s_n\}$ is a Cauchy sequence in $X$, then I want to make a series. For this I do not have any idea. Please help me.

REPLY [5 votes]: Let $x_n$ be a Cauchy sequence. The goal is to show that there exists $x \in X$ such that $x_n \to x$. 
Note that if we can construct a subsequence $x_{n_k}$ of $x_n$ that converges to $x$ then also $x_n\to x$ since if $x_{n_k}$ is $\varepsilon$-close to $x$ then so are all points $x_n$ with $n > n_k$.
Let's construct $x_{n_k}$. Start with $\varepsilon = {1\over 2}$. Then there exists $N_1$ such that for $n,m\ge N_1$:
$$ \|x_n - x_m \|<{1\over 2}$$
Let $x_{n_1}=x_{N_1}$. Proceed in a similar fashion so that for $n,m \ge N_k$ we have
$$ \|x_n - x_m \|<{1\over 2^k}$$
and let $x_{n_k}=x_{N_k}$.
Then 
$$ \sum_{k=1}^\infty \|x_{n_{k+1}}-x_{n_k}\| <\infty$$
hence by assumption $\sum_{k=1}^\infty (x_{n_{k+1}}-x_{n_k})$ converges. Let $S=\sum_{k=1}^\infty (x_{n_{k+1}}-x_{n_k})$ denote its limit. Note that 
$$\sum_{k=1}^K (x_{n_{k+1}}-x_{n_k}) = x_{n_K}-x_{n_1}$$ 
By the algebraic limit theorem we know that
$$ \lim_{K \to \infty} x_{n_K} = S + x_{n_1}$$
hence $x_{n_k}$ is a convergent subsequence as desired.