Let $A(x)$ be a Dirichelet generating function given by: $A(x) = \frac{a_1}{1^x}+\frac{a_2}{2^x}+\frac{a_3}{3^x}+...=\sum_{n=1}^{\infty}\frac{a_n}{n^x}$.
Given the Dirichelet generating functions $A(x)$, $B(x)$, and $C(x)$, where $C(x) = A(x) B(x)$, find a formula for $c_n$ in terms of the $a_i$ s and $b_j$ s.
--Attempt--
Find the first few terms of $c_n$ as:
$(\frac{a_1}{1^x})(\frac{b_1}{1^x})=\frac{a_1b_1}{1^x}$ and so the coefficient of $\frac{1}{1^x}$, that is, $c_1$, is $a_1b_1$
$(\frac{a_1}{1^x}+\frac{a_2}{2^x})(\frac{b_1}{1^x}+\frac{b_2}{2^x})=\frac{a_1b_1}{1^x}+\frac{a_1b_2+a_2b_1}{2^x}$ and so the coefficient of $\frac{1}{2^x}$ is $a_1b_2+a_2b_1$
$(\frac{a_1}{1^x}+\frac{a_2}{2^x}+\frac{a_3}{3^x})(\frac{b_1}{1^x}+\frac{b_2}{2^x}+\frac{b_3}{3^x})=\frac{a_1b_1}{1^x}+\frac{a_1b_2+a_2b_1}{2^x}+\frac{a_1b_3}{3^x}+\frac{a_2b_2}{4^x}+\frac{a_3b_1}{3^x}$ and so the coefficient of $\frac{1}{3^x}$ is $a_1b_3+\frac{4}{3}^{-x}a_2b_2+a_3b_1$
I'm confident in $c_1$ and $c_2$, but feel that I'm missing something for $c_3$ and on. Any push in the right direction would be much appreciated!
We obtain \begin{align*} \sum_{n=1}^\infty \frac{\color{blue}{c_n}}{n^x} &=\left(\sum_{k=1}^\infty\frac{a^k}{k^x}\right)\left(\sum_{l=1}^\infty\frac{b^l}{l^x}\right)\\ &=\sum_{n=1}^\infty\left(\sum_{{k\cdot l=n}\atop{k,l\geq 1}}\frac{a_k}{k^x}\,\frac{b_l}{l^x}\right)\\ &=\sum_{n=1}^\infty \left(\sum_{{k=1}\atop{k|n}}^n\frac{a_k}{k^x}\,\frac{b_{n/k}}{\left(n/k\right)^x}\right)\\ &=\sum_{n=1}^\infty \color{blue}{\left(\cdots\right)}\frac{1}{n^x}\\ \end{align*}
Do you see how to do the final step(s) and derive a formula for $\color{blue}{c_n}$? You might also want to calculate $c_3$ and $c_4$ from it to better see what's going on.