Right implication is clear. Let $(t_n)_n$ be such that $t_n \rightarrow 0$. Let $N$ be such that for $n > N$, $|t_n| < 1$. Then $\sum_n s_n t_n \leq \sum_n |s_n| |t_n| \leq \sum_n^N |s_n| |t_n| + \sum_{n=N} |s_n| < \infty$
For the second implication here is my progress so far:
We have that $\sum_n s_n t_n$ converges for each $t_n$ convergent to $0$, now we have to prove $(s_n)_n \in \ell^1$.
Consider the application $J: \ell^\infty \rightarrow (\ell^\infty)^*$, $s \mapsto Js$ where $Js$ is defined as $Js((t_n)_n)= \sum_n s_n t_n$. Clearly $Js$ is linear. My idea is that if we can prove that $Js$ is continuous in a neighbour of $0$ then $Js$ is bounded so we have $\sum_n |s_n| =Js(sign(s_n)) \leq ||Js || < \infty$.
Im having some trouble in proving countinuity at $0$ of $Js$ using only that $Js(t_n) < \infty$ if $t_n$ converges to $0$. I have tried to show that $Js$ is graph-closed, i.e. $(s, Js) \subset \ell^\infty \oplus \ell^\infty$ is closed, hence continuous but with not much success.
Note that if $\sum_k s_k$ is not absolutely convergent, then $\sum_{k \ge n} |s_k| = \infty$ for all $n$.
Let $n_0 = 0$, and choose $n_i$ such that $\sum_{n_i < k \le n_{i+1}} |s_k| \ge 2^i$.
Define $t_k = {1 \over 2^i}\operatorname{sgn} s_k$ for $n_i < k \le n_{i+1}$.
Then $\sum_{n_i < k \le n_{i+1}} t_k s_k \ge 1$.