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?



Summation of bounded sequences
Your condition is equivalent to the convergence in $\ell^1$ of $b\_n=(b\_{nj})\_{j\ge 1}$.


First of all, we indeed must insist that $b\_n=(b\_{nj})\_{j\ge 1}\in\ell^1$ for each fixed $n$, or otherwise we couldn't even define $\sum b\_{nj}s\_j$ for general bounded $s$.


Recall that $(\ell^1)^\*=\ell^{\infty}$, and observe that if $(b\_n,s)\to L(s)$ for all $s\in\ell^{\infty}$, then $\|b\_n\|\_1$ must be bounded by the uniform boundedness principle. (Or a direct argument, by hand, could be given, and you could check an earlier version of this answer for this if interested.)


This implies that the pointwise limit $b\_j=\lim b\_{nj}$ defines an $\ell^1$ sequence.
We want to show that $\|b\_n-b\|\_1\to 0$. Originally, I thought this would just follow from the well known fact that weakly convergent sequences in $\ell^1$ are norm convergent. However, it's not so clear at this point that $b\_n\to b$ weakly.


Instead, we make use of the argument from the [usual proof](https://math.stackexchange.com/questions/2208898/schurs-theorem-in-ell1-weak-convergence-of-x-n-is-the-same-as-convergenc) of the fact just quoted. It suffices to show that $b\_n$ is uniformly summable in the sense that for every $\epsilon>0$, there exists $J\ge 1$ so that $\sum\_{j\ge J}|b\_{nj}|<\epsilon$ for all $n$. Suppose this were not the case, so there is an $\epsilon>0$ such that given any $J\ge 1$, we can find an $n\_J$ so that $\sum\_{j\ge J}|b\_{n\_Jj}|\ge 3\epsilon$. In fact, we can then also find *arbitrarily large* $n\_J$'s with this property.


Let's use this now to pick an $n\_1$ and an interval $I\_1$ such that: (1) $\sum\_{j\in I\_1}|b\_{n\_1j}|\ge 2\epsilon$; (2)$\sum\_{j\in I\_1}|b\_j|<2^{-1}\epsilon/10$; (3) $\sum\_{j>\max I\_1}|b\_{n\_1j}|<\epsilon$.


Now we take $s\_j$'s of absolute value $1$ on $I\_1$ such that $b\_{n\_1j}s\_j=|b\_{n\_1j}|$. Note that this ensures that $\sum\_{j\ge 1} b\_{n\_1j}s\_j$ will be close to a positive number of size $\ge\epsilon$, and this is true for any choice of $s\_j$'s of absolute value $1$ for larger $j$'s at later stages.


Now we just continue in this style: we pick an index $n\_2>n\_1$ and an interval $I\_2$. Note that if $n\_2$ is sufficiently large, then what we did on $I\_1$ will not have much of an effect on $\sum b\_{n\_2j}s\_j$ because for $j\in I\_1$, we will
have $b\_{n\_2j}$ very close to $b\_j$, and then property (2) from above strikes. We now arrange matters so that $\sum b\_{n\_2j}s\_j$ is close to a *negative* number $\ge\epsilon$ in absolute value.


Of course this oscillating pattern is incompatible with convergence, so it follows that $b\_n$ is uniformly summable, and thus $b\_n\to b$ in $\ell^1$, as claimed.


