If $\sum_n \|x\|< \infty$, how to show that $\sum x_n$ is convergent in the Hilbert space $H$. Let $\{x_n\}$ be a sequence in a Hilbert space $H$. If $\sum_n \|x\|< \infty$, how to show that $\sum x_n$ is convergent in $H$?
There is no doubt that $x_n \rightarrow 0$ as $n \rightarrow \infty$ (right?) since we have that
\begin{align*}
\sum_n \|x\|< \infty &\iff
\|x_n\|\rightarrow 0 \text{ as } n \rightarrow \infty \\
&\iff \langle x_n,x_n\rangle \rightarrow 0 \text{ as } n \rightarrow \infty \\
&\iff x_n \rightarrow 0 \text{ as } n\rightarrow \infty
\end{align*}
(it is a property of the inner product that $\langle x,x\rangle=0 \implies x=0$)
What to do next? Can I use the completeness of $H$ somehow?
 A: The series $\sum_n ||x_n||$ is convergent if and only if the partial sum $\sum_{k=1}^n||x_k||$ is a cauchy sequence hence
$$\forall \epsilon >0\ \exists\ n_0\quad \forall p\ge q\ge n_0,\  ||\sum_{k=p}^q x_k||\le\sum_{k=p}^q||x_k||<\epsilon $$
and hence  $\sum_{k=1}^n x_k$ is a Cauchy sequence in a complete space. Conclude.
A: Try to show that $y_k = \sum_{i=1}^k x_i$ is a Cauchy sequence.
A: Yes, you can use completeness; a normed vector space X is complete iff every absolutely-summable --meaning $\Sigma ||x_n||$ converges-- is summable, i.e., if $\Sigma x_n$ converges in norm, i.e., if there is an $x$ in $X$ with $||x- \Sigma x_i||\rightarrow 0$. Basically, if $\Sigma ||x_n||< \infty$ , there is a pos. integer N so that $\Sigma _{n=N}^{\infty}||x_n||< \epsilon$. Then, for $j,k > N $, we have, for the partial sums $S_n, S_m$, that $$ ||S_n-Sm||=||\Sigma_ {x=m}^{n}x_k||\leq \Sigma_{k=m}^{\infty} ||x_k||< \epsilon$$ , so you have a Cauchy sum in $X$ , which then converges by 
completeness of $X$.
