Let $(X,\|\cdot\|)$ be a normed vector space. Show that $X $ is a Banach space if and only if for every succession $\{x_n\}_{n\in \mathbb N} $ such that $\sum_{n=1}^\infty ||x_n|| \lt \infty $, we have that $\sum_{n=1}^\infty x_n $ converges to an element of $X $.
If $X $ is Banach, it's clear that the statement above is true for every succession $\{x_n\}_{n\in \mathbb N} $ such that $\sum_{n=1}^\infty ||x_n|| \lt \infty $, because this implies that $\{S_N\}_N=\sum_{n=1}^N x_n $ is a Cauchy succession in $X $.
However, in the other sense, I don't understand what to do: if I take a Cauchy succession $\{x_n\}_n$ in $X $, I could show that it always converges by showing that $\sum_{n=1}^\infty (x_n - x_{n-1}) $ converges to some point of $X $. However it is not true in general that $\sum_{n=1}^\infty ||x_n-x_{n-1}|| \lt \infty $, even for a Cauchy succession. Do you have suggestions? Thanks
Here's the argument used in Folland; it's fairly standard. You'll see a fairly similar subsequence construction in showing that $L^1$ is complete, for example.
Take a Cauchy sequence $(x_n)$. Since it's Cauchy, for each $j$, you can find $n_j$ (which WLOG are increasing) so that $\|x_n-x_m\|<2^{-j}$ for $m,n\geq n_j$. If you take $y_1=x_{n_1}$ and $y_j=x_{n_j}-x_{n_{j-1}}$ for $j>1,$ then you can directly see that $$\sum\limits_{j=1}^k y_j=x_{n_k}$$ and $$\sum\limits_{j=1}^\infty \|y_j\|\leq \|y_1\|+\sum\limits_{j=1}^\infty 2^{-j}=\|y_1\|+1<\infty.$$ Using our assumptions, this means that $$\sum\limits_{j=1}^\infty y_j=\lim_k x_{n_k}$$ converges. But, if a Cauchy sequence has a convergent subsequence, then it converges to the same limit. So, $(x_n)$ converges, which implies that $X$ is Banach.