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.