TITLE: A series in a Hilbert space that converges unconditionally but only in weak topology
QUESTION [6 upvotes]: Let $H$ be a Hilbert space. I'm looking for an example of a series $\sum x_n$ with the following two properties: 

$\sum x_n$ does not converge, i.e., partial sums do not have a limit in the norm topology of $H$;
$\sum x_n$ converges unconditionally in the weak topology, that is for every bijection $\sigma:\mathbb{N}\to\mathbb{N}$ the partial sums of the rearranged series $\sum x_{\sigma(n)}$ converge weakly. 


Progress so far: considered some weakly but not strongly convergent sequences, and corresponding series. In all cases the weak convergence turned out to be conditional. For example: let $e_n$ be an orthonormal basis, and consider $x_1=e_1$, $x_n = e_n - e_{n-1}$ for $n>1$. Then the partial sums of $\sum x_n$ are precisely the vectors $e_n$ which converge to zero weakly but not in the norm. However, this weak convergence is conditional: if we take only even-numbered elements
$$x_2+x_4+x_6+\dots = (e_2-e_1) + (e_4-e_3) + (e_6-e_4)+\dots$$
the norm of partial sums tends to $\infty$; and including an odd-numbered $x_n$ once in a million terms will not change that. Since a weakly convergent sequence must be bounded, the rearranged series does not converge weakly.

REPLY [2 votes]: I'll expand on the reference given by Kavi Rama Murthy. In Normed Linear Spaces, Chapter IV section 1, M.M. Day considers four definitions that look like unconditional convergence of a series $\sum x_n$ (in a locally convex space $L$): 
(B) Reordered convergent: there exists $x$ such that $\sum x_{\pi(n)} =x$ for every bijection $\pi:\mathbb{N}\to\mathbb{N}$
(C) Unordered convergent: there exists $x$ such that for every neighborhood $U$ of $x$ there exists a finite set $E\subset \mathbb{N}$ with the property that $\sum_{n\in F}x_n\in U$ for every finite set $F$ containing $E$.
(D) Subseries convergent: for every increasing sequence of integers $(n_k)$ the series $\sum x_{n_k}$ converges.
(E) Bounded-multiplier convergent: for every bounded sequence of scalars $(a_n)$, the series $\sum a_n x_n$ converges.
In every LCS one has $(E)\implies (D)\implies (C) \iff (B)$. Moreover, if one replaces the requirement of convergence in (B, C, D, E) by the Cauchy property, then all the resulting conditions are equivalent. So, if $L$ is sequentially complete, then (B, C, D, E) are equivalent in $L$. 
Theorem 1 (page 80): Let $B$ be a Banach space. If $\sum x_n$ is subseries convergent in the weak topology of $B$, then it is subseries convergent in the norm topology of $B$. 
Therefore, if $B$ is weakly sequentially complete, then there are no examples of the kind I was looking for. All reflexive spaces and all $L^1$ spaces are weakly sequentially complete. (Reference).