The question asks: if $\sum a_n$ converges, $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges.
My proof goes as follows:
Let $\varepsilon>0$, and let $S_k$ denote $k$-th partial sum of $\sum a_nb_n$. Now, $\{b_n\}$ is monotonic and bounded, so it converges, say to $b$.
Since $\{b_n\}$ converges, we know that $\|b_n\| < \|b\|+1$, for large enough $n$. Also, $\sum a_n$ converges, thus $\|\sum_{j+1}^{k} a_n\| < \frac{\varepsilon}{\|b\|+1}$ for large enough $k,j$.
Therefore $\|S_k-S_j\| = \|\sum_{j+1}^{k} a_nb_n\| < (\|b\|+1)\|\sum_{j+1}^{k} a_n\|<(\|b\|+1)\times\frac{\varepsilon}{\|b\|+1}=\varepsilon,$ for large enough $k,j$.
Thus, $\sum a_nb_n$ is Cauchy, hence convergent.
Now, is my proof correct? Thanks for the help.