TITLE: Summation of bounded sequences
QUESTION [6 upvotes]: Let $b_{n,j}\in \mathbb{C}$ for each $n,j\in \mathbb{N}$. I was wondering if there is some characterization of those $b_{n,j}$ such that for all bounded sequences $s_j\in \mathbb{C}, j\in \mathbb{N}$, we have that
$$\sigma_n := \sum_{j=1}^{+\infty} b_{n,j} s_j$$
converges. Thus we want that the sequence $(\sigma_n)_n$ converges for all bounded sequences.
Note that I do not ask that $s_j\rightarrow s$ implies $\sigma_n\rightarrow s$ as is usual in summability methods.
So can we characterize exactly which $b_{n,j}$ satisfy this?

REPLY [5 votes]: A comment on  Christian Remling's answer. As proved there, we know that $(b_n)$ is a  sequence in $\ell_1$, and  by assumption we have that $\langle b_n, s\rangle$ converges in $\mathbb{C}$ for any $s\in\ell_\infty$. This defines an element  $\tilde b$ of $\ell_\infty^*=\ell_1^{**}$ such that $b_n\to \tilde b$ in the weak$^*$ topology of $\ell_\infty^*$, that is $\sigma(\ell_\infty^*,\ell_\infty)$. The key point is that $\ell_1$ is weakly sequentially complete (essentially proven in C.R. answer), so that $\tilde b$ actually is in $\ell_1$ (not just $\ell_1^{**}$) and the convergence is also in the  $\ell_1$ norm as said. (Reference for $L^1$ spaces are weakly sequentially complete: Dunford-Schwarz, Linear Operators, Part I, IV.8.6).
rmk. In fact Schur's theorem on the equivalence of strong and weak convergence in $\ell_1$, that is the fact that a weak  convergent  sequence  is also a  norm convergent sequence, extends to a more complete statement : a weak Cauchy sequence is also  a norm Cauchy sequence (whence the weak sequential completeness). Indeed, if $b_n$ is a weak-Cauchy sequence, for any pair of diverging sequences of integers $n_k$ and $m_k$, one has $b_{n_k}- b_{m_k}\to0$ weakly as $k\to0$, therefore also $\|b_{n_k}- b_{m_k}\|\to0$.