Vtyjkyuk

Show that

A Normed Linear Space $X$ is a Banach Space iff every absolutely convergent series is convergent.

My try:

Let $X$ is a Banach Space .Let $sum x_n$ be an absolutely convergent series .Consider $s_n=sum_i=1^nx_i$. Now $sum |x_n|<infty implies exists N$ such that $sum_i=N^ infty |x_i|<epsilon$ for any $epsilon>0$

Then $|s_n-s_m|le sum _i=m+1^n |x_i|<epsilon forall n,m>N$

So $s_n$ is Cauchy in $X$ and hence converges $s_nto s$ (say).

Thus $sum x_i$ converges.

Conversely, let $x_n$ be a Cauchy Sequence in $X$. Here I can't proceed how to use the given fact.

Any help will be great.

For the converse argument; let $X$ be a normed linear space in which every absolutely convergent series converges, and suppose that $x_n$
is a Cauchy sequence.

For each $k in mathbbN$, choose $n_k$ such that $||(x_m − x_n)|| < 2^−k$ for $m, n geq n_k$. In particular, $||x_n_k+1−x_n_k||< 2^-k$. If we define $y_1 = x_n_1$ and $y_k+1 = x_n_k+1 − x_n_k$
for $k geq 1$, it follows that $sum ||y_n|| ≤ ||x_n_1 || + 1$
i. e., ($y_n$) is absolutely convergent, and hence convergent.

Hint: show that there is a subsequence $y_n$ such that $|y_n - y_n+1| < 2^-n$.