If every absolutely convergent series is convergent then $X$ is Banach

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.

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.

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