(This question has been on [math.SE](http://math.stackexchange.com/q/734052/57159) for over a week and has not gotten any answers.)

Let $G$ be a (T$_0$) topological abelian group, and let $0$ be its identity element.

Assume that for all index sets $I$ and all functions $f\colon I\to G$, if

$\big[$for each neighborhood $U$ of $0$, $f(i)\in U$ for all but finitely many $i$ $\big]$

then $\: $[$\displaystyle\sum_{i\in I}\hspace{.03 in}f(i)$](http://en.wikipedia.org/wiki/Series_%28mathematics%29#Abelian_topological_groups)$ \:$ exists.

Does it follow that every neighborhood of $0$ contains an open subgroup
$H$ of $G$?