Completeness in normed space

A normed space X is complete if and only if every absolutely convergent series converges

If a normed space is complete, then every absolutely convergent series converges.

Suppose that $ $is a absolutely convergent series.(i.e.$\sum_{n=1}^{\infty}||x_n||<\infty$ and denote

$$s_n=\sum_{i=1}^{n}x_i$$

Because $\ $ is a absolutely convergent series, there exists $k>0$ such that $$\sum_{n=k+1}^{\infty}x_n <\epsilon$$

$$||s_m-s_n||=|| \sum_{i=n+1}^{m} x_i||\le \sum_{i=n+1}^{m}||x_i||$$

if we let $n>k$, we can prove that $\ $ is a Cauchy series, then converges.(X is complete)

We need to prove if every absolutely convergent series in a normed space converges, then the normed space X is complete.

Let$\ $ be an Cauchy sequence in X and therefore $\forall \epsilon>0, \exists p_k\in \mathbb{N}, \forall m,n>p_k$, such that

$$ || x_m-x_n ||\le 2^{-k} $$

WLOG,we can assume $\ $ is strictly increasing.
Then the series $\sum_{k=1}^{\infty} (x_{p_{k+1}}-x_)$ is absolutely convergent and therefore, convergent and therefore, the sequence

$$ x_{ p_k }=x_{p_1}+(x_{p_2}-x_{p_1})+(x_{p_3}-x_{p_2})+\dots+(x_{p_k}-x_{p_{k-1}}) $$

converges to an element $x \in X$
Then
$$ || x_n-x ||\le|| x_n-x_{p_k} ||+|| x_{p_k} -x||\to 0 $$
Q.E.D