 Solution

The series ${\sum }_{n\ge 1}{x}_n$ converges in $X$ when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ converge in norm to an element of $X$. It is absolutely convergent when ${\sum }_n\Vert x_n\Vert <\infty$.

If $X$ is complete, thenfor 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 $\Vert z_{n_{k+1}}-z_{n_k}\Vert \le 2^{-k}$. The seriesis 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 $\Vert z_n-z\Vert$ by $\Vert z_n-z_{n_k}\Vert +\Vert z_{n_k}-z\Vert$. Hence