If every absolutely convergent series is convergent then $X$ is Banach 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 $s_n\to s$ (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.
 A: A normed space is complete if and only if every absolutely convergent series converges.
$\implies$ 
We will prove this by proving absolutely convergent series is a Cauchy series.
We define a absolutely convergent series. Suppose $x_n\in E$ and $\sum_{n=1}^\infty||x_n||<\infty$ and denote
$$
s_n=\sum_{k=1}^nx_k
$$
Because the sequence of partial sums converges in E, for every $\epsilon >0 $, there exists $k>0$ such that
$$
\sum_{n=k+1}^\infty ||x_n||<\epsilon
$$
To show $(s_n)$ is a Cauchy sequence,  $\forall \epsilon>0,\exists M,\forall m,n>M$ such that 
$$
||s_m-s_n||=||x_{n+1}+x_{n+2}+\dotsm+x_m||\le \sum_{r=n+1}^\infty ||x_r||<\epsilon
$$
(without loss of generality, we assume $m>n$)
Since E is complete, $s_n$ converges.
$\Longleftarrow$ 
We need to prove if every absolutely convergent series in a normed space converges, then the normed space is complete.
Let $(x_n)$ be an Cauchy sequence in E and therefore $\forall \epsilon>0,\exists p_k\in N,\forall m,n>p_k$ such that 
$$
||x_m-x_n||<2^{-k}
$$
without loss of generality, we can assume $(p_k)$ is strictly increasing.
Then the series $\sum_{k=1}^\infty (x_{p_{k+1}}-x_{p_k})$ is absolutely convergent and therefore, convergent and therefore, the sequence
$$
x_{p_k}=x_{p_1}+(x_{p_2}-x_{p_1})+(x_{p_3}-x_{p_2})+\dotsm+(x_{p_k}-x_{p_{k-1}})
$$
converges to an element $x\in E$ 
Then 
$$
||x_n-x||\le ||x_n-x_{p_n}||+||x_{p_n}-x||\rightarrow 0
$$
Q.E.D.
A: Hint: show that there is a subsequence $y_n$ such that $\|y_n - y_{n+1}\| < 2^{-n}$.
A: 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 $k \in \mathbb{N}$, 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.
