= Solution
The <series> $\sum_{n\geq1}x_n$ converges in $X$ when its <partial sums> $s_N=\sum_{n=1}^Nx_n$ converge in <norm> to an element of $X$. It is absolutely convergent when $\sum_n\|x_n\|<\infty$.
If $X$ is complete, then
$$
\|s_M-s_N\|\leq\sum_{n=N+1}^M\|x_n\|\longrightarrow0
$$
for an absolutely convergent <series>. Its <partial sums> are Cauchy and therefore converge.
Conversely suppose every absolutely convergent <series> converges. Given a <Cauchy sequence> $(z_n)$, choose a subsequence $(z_{n_k})$ with $\|z_{n_{k+1}}-z_{n_k}\|\leq2^{-k}$. The <series>
$$
z_{n_1}+\sum_{k=1}^{\infty}(z_{n_{k+1}}-z_{n_k})
$$
is absolutely convergent. Its <partial sums> are precisely the subsequence, which therefore converges to some $z\in X$. A <Cauchy sequence> with a convergent subsequence converges to that same limit: estimate $\|z_n-z\|$ by $\|z_n-z_{n_k}\|+\|z_{n_k}-z\|$. Hence
$$
\boxed{X\text{ is Banach}\iff\text{every absolutely convergent series in }X\text{ converges}.}
$$
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