# Banach space and sequence

Let $(X,\left\Vert \cdot \right\Vert)$ be a normed space. Show that $X$ is Banach space (under the given norm) if and only if the sum $\Sigma_{n=1}^{\infty}x_n$ converges in $X$ for any sequence $(x_n)_{n=1}^{\infty}$ of vectors in $X$ with $\Sigma_{n=1}^{\infty}\left\Vert x_n \right\Vert<\infty$.

To prove $\implies$, $X$ is a Banach space is given, and I need to show that the sum converges.

But I am a bit confused because I thought $\Sigma_{n=1}^{\infty}x_n$ is finite if $\Sigma_{n=1}^{\infty}\left\Vert x_n \right\Vert<\infty$ is given and therefore it converges since the sum to infinity is finite. Please correct me if I am wrong.

We really have to use the fact that the space is a Banach space. The fact that $\sum_{n=1}^{\infty}\left\lVert x_n\right\rVert$ do not guarantee in general the convergence of the sequence $\left(\sum_{n=1}^Nx_n\right)_{N\geqslant 1}$. For example, let $c_{00}$ be the space of sequence of real number such that only finitely many terms may not vanish. Endow this space with the supremum normand let $x_n$ be the element of $c_{00}$ whose $n$-th term is $2^{-n}$ and all the others are zero. The norm of $x_n$ is $2^{-n}$but there is no convergence in $c_{00}$.
Set $S_{n} = \sum_{k = 1}^{n}x_{n}$ for $n \geq 1$. Then we need to show that $\{S_{n}\}_{n =1}^{\infty}$ has a limit. Since $X$ is a Banach space it will suffice to show that it is Cauchy. If $n > m$, then $||S_{n} - S_{m}|| = || \sum_{k = m+1}^{n} x_{k}|| \leq \sum_{k = m+1}^{n} ||x_{k}||$. Since $\sum_{k = 1}^{\infty}||x_{k}|| < \infty$, $\sum_{k = m+1}^{n}||x_{k}|| \rightarrow 0$ as $m,n \rightarrow \infty$ and hence $\{S_{n}\}_{n = 1}^{\infty}$ is Cauchy.
• As for the other direction, if I have a Cauchy sequence, I know I can choose a subsequence such that $\Sigma_{k=1}^{\infty}\left\Vert x_{n_{k+1}}-x_{n_k} \right\Vert < \infty$ and thus $\Sigma_{k=1}^{\infty}(x_{n_{k+1}}-x_{n_k})$ converges. But how do I proceed from here? – LanaDR May 6 '18 at 14:07