A series in a Hilbert space that converges unconditionally but only in weak topology 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.
 A: 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).
A: There is no such example. Theorem 1, p. 80 Of Normed Linear Spaces by M. M. Day (Third edition)  implies that if the series unconditionally convergent in the weak topology then $\sum x_n$ converges in the norm. 
