Proving that $X$ is a Banach space iff convergence of $\sum\|x_n\|$ implies convergence of $\sum x_n$ 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.  
 A: 1) Choose a subsequence $\{s_{n_k}\}$ of $\{s_n\}$ such that
$$
\|s_{n_{k+1}} - s_{n_k}\| < 2^{-k} \quad\forall k
$$
and show that it converges using the hypothesis.
2) Show that if a subsequence of a Cauchy sequence converges, then the entire sequence converges.
A: 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.
A: Let $\{x_n\}$ be a Cauchy sequence. We need to show that it converges. 
It suffices to show that $\{x_n\}$ possesses a converging subsequence.
As it is Cauchy, for every $\varepsilon>0$, and in particular for $\varepsilon=2^{-k}$, there exists an $N=N(k)$, such that, $m,n\ge N(k)$ implies that
$$
\|x_m-x_n\|<2^{-k}.
$$
The $N(k)$'s can be chosen to form a strictly increasing sequence.
Now we shall show that the subsequence $y_k=x_{N(k)}$ converges. 
Note that $\|y_k-y_{k-1}\|<2^{{-k}}$.
In particular, 
$$
y_k=y_1+(y_2-y_1)+(y_3-y_2)+\cdots+(y_k-y_{k-1})=z_1+z_2+\cdots+z_k,
$$
and for $z_k$ we have that
$$
\sum \|z_k\|\le \|y_1\|+\sum_{k=1}^\infty 2^{-k}=\|y_1\|+\frac{1}{2}<\infty,
$$
and thus $y_k$ converges. Let $y_k\to y$.
It is left to you to show that $x_n$ converges as well to $y$.
