Let me give a proof.
From the assumption, of course
$\sum a_n$ converges. And also
$\sum a_nb_n$ converges for all $\lim b_n=0$. Note here is $0$ since one can replace $b$ by $b-1$ to get convergence.
Now assume to the contrary that $\sum a_n$ does not converge absolutely.
Take $A_n=\max\{a_n,0\}$, the non-negative part and $B_n=-\min\{a_n,0\}$, the non-positive part. Then $\sum|a_n|=\sum A_n+\sum B_n=\infty$. By symmetry, we may assume that $\sum A_n$ is infinity (or just take the non-positive part instead). So we can find a increasing sequence of interger $n_k$ inductively such that $n_0=1$ and $$\sum_{n_k+1}^{n_{k+1}}A_n>k.$$
Now take
$$b_n =
\begin{cases}
\frac{1}{k} & \text{ if } n_k<n\leq n_{k+1} \text{ and } A_n\neq 0;\\
0 &\text{ if } A_n=0.
\end{cases}$$
Then one can see that $\lim b_n=0$ but
$$\sum a_nb_n=\sum A_nb_n=\sum_k\sum_{n_k+1}^{n_{k+1}}A_n\cdot\frac{1}{k}>\sum_k1=\infty.$$ A contradiction.