$\{ a_n \} ,\{b_n\}$ : bounded sequences
I have to prove this proposition;
For $s\in \mathbb{C} \,\, (\text{Re}(s)>1),$
$
\left( \sum_{k=1}^ {\infty} \dfrac{a_k}{k^s} \right)
\left(\sum_{l=1}^{\infty} \dfrac{b_l}{l^s}\right)
\text{ and } \sum_{m=1}^{\infty} \dfrac{c_m}{m^s}
\quad (c_m:=\sum_{k,l\geqq 1, kl=m} a_k b_l)$
converge absolutely. Furthermore,
$\left( \sum_{k=1}^ {\infty} \dfrac{a_k}{k^s} \right)
\left(\sum_{l=1}^{\infty} \dfrac{b_l}{l^s}\right)
=\sum_{m=1}^{\infty} \dfrac{c_m}{m^s}
$
My attempt is as following.
Because $\{a_n\}$ and $\{b_n\}$ are bounded, there exist $K, M>0$ such that $|a_n|\leqq K, |b_n|\leqq M$ for all $n$.
Therefore,
$ \sum_{k=1}^ {\infty} \Bigg|\dfrac{a_k}{k^s}\Bigg| \leqq K\sum_{k=1}^{\infty} \dfrac{1}{k^s}, \,\, \sum_{l=1}^ {\infty} \Bigg|\dfrac{b_l}{l^s}\Bigg| \leqq M\sum_{l=1}^{\infty} \dfrac{1}{l^s}$. And because Re$(s)>1$, both convergent absolutely. Thus, $ \left( \sum_{k=1}^ {\infty} \dfrac{a_k}{k^s} \right) \left(\sum_{l=1}^{\infty} \dfrac{b_l}{l^s}\right)$ convergent absolutely.
But I cannot prove $ \sum_{m=1}^{\infty} \dfrac{c_m}{m^s} $ convergents absolutely and $\left( \sum_{k=1}^ {\infty} \dfrac{a_k}{k^s} \right) \left(\sum_{l=1}^{\infty} \dfrac{b_l}{l^s}\right) =\sum_{m=1}^{\infty} \dfrac{c_m}{m^s}.$
For the convergence of $\sum_{m=1}^{\infty} \dfrac{c_m}{m^s}$, I tried \begin{align} \sum_{m=1}^{\infty} \Bigg| \dfrac{c_m}{m^s} \Bigg| &=\sum_{m=1}^{\infty} \Bigg( \dfrac{1}{m^s} \Bigg|\sum_{k,l\geqq 1, kl=m} a_k b_l \Bigg| \Bigg)\\ &\leqq \sum_{m=1}^{\infty} \Bigg( \dfrac{1}{m^s} \sum_{k,l\geqq 1, kl=m}\Bigg| a_k b_l \Bigg|\Bigg) \\ &\leqq \sum_{m=1}^{\infty} \Bigg( \dfrac{1}{m^s} \sum_{k,l\geqq 1, kl=m}KM \Bigg). \\ \end{align} But I cannot proceed. And I have no idea to prove $\left( \sum_{k=1}^ {\infty} \dfrac{a_k}{k^s} \right) \left(\sum_{l=1}^{\infty} \dfrac{b_l}{l^s}\right) =\sum_{m=1}^{\infty} \dfrac{c_m}{m^s}$.
I would like you to give me some ideas.