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.
Best Answer
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 $y$.