banach spaces have all absolutely summable series are summable

A normed vector space $V$ is a Banach space if and only if every absolutely summable series is summable.

$(⟹)$ Suppose $V$ is a Banach space. If $\sum _n{v}_n$ is absolutely summable, then the sequence of partial sums is a Cauchy sequence in $V$ (see the previous theorem). Thus ${\left\{\sum _{n=1}^m{v}_n\right\}}_m$ converges in $V$. Thus by definition, the series is summable.

$(⟸)$ Suppose every absolutely summable series is summable. Let $\{v_n\}$ be a Cauchy sequence in $V$. We show that a subsequence of $\{v_n\}$ converges in $V$. Then $\{v_n{\}}_m$ converges by metric space theory.

$\{v_n{\}}_m$ is Cauchy $⟹$ for all $k\in \mathbb{N}$, there exists $N_k\in \mathbb{N}$
such that for all $n,m\ge N_k$, we have

$$||v_n-v_m||<\frac{1}{2^k}$$

(we choose this expression because it is summable). Now, define

$$n_k=N_1+N_2+\cdots +N_k$$

Then $n_1<n_2<\dots$ and for all $k$, we have $n_k\ge N_k$. Thus, for all $k\in \mathbb{N}$, we have

$$||v_{n_{k+1}}-v_{n_k}||<\frac{1}{2^k}$$

Thus

$$\begin{array}{rl}& \sum _k(v_{n_{k+1}}-v_{n_k})\text{ is absolutely summable}\\ ⟹& \sum _k(v_{n_{k+1}}-v_{n_k})\text{ is summable}\\ ⟹& {\{\sum _{k=1}^m(v_{n_{k+1}}-v_{n_k})\}}_{m=1}^{\mathrm{\infty }}=\{v_{n_{m+1}-v_{n_1}}{\}}_{m=1}^{\mathrm{\infty }}\text{ converges}\end{array}$$

Thus the (sub)sequence is $\{v_{n_{m+1}}{\}}_{m=1}^{\mathrm{\infty }}$ converges in $V$. Thus $V$ is Banach. $◼$