Product of two absolutely convergent Dirichlet series We have$$(f * g)(n) = \sum_{d \mid n} f(d)g(n/d).$$How do I see that if the two Dirichlet series$$F(s) = \sum_{n =1}^\infty f(n)n^{-s},\text{ }G(s) = \sum_{n=1}^\infty g(n)n^{-s}$$converge absolutely for $\text{Re}(s) > \sigma_0$, then in the same half-plane, the following equations hold:$$F(s)G(s) = \left( \sum_{n=1}^\infty f(n)n^{-s}\right)\left( \sum_{n=1}^\infty g(n)n^{-s}\right) = \sum_{n=1}^\infty (f * g)(n)n^{-s}?$$
 A: $$
\sum_{n\geq 1}\frac{(f*g)(n)}{n^s} =
\sum_{n\geq 1}\frac{1}{n^s}\sum_{d\mid n}f(d)g(n/d)
= \sum_{k\geq 1} \sum_{d\geq 1}\frac1{(kd)^s}f(d)g(k) \;.
$$
A: We need a trick of change of variablies. Notice that 
\begin{align}
F(s)G(s)&=\sum_{n=1}^\infty \frac{f(n)}{n^s}\sum_{m=1}^\infty \frac{g(m)}{m^s}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{f(n)g(m)}{(mn)^s}
\end{align}
Now we make a change of variables and use indicator functions to remove the constaint. For fixed $n$, let $m'=mn$, then we get
\begin{align}
F(s)G(s)&=\sum_{n=1}^\infty \sum_{m'\geq 1:n\ \text{divides}\ m'}\frac{f(n)g(m'/n)}{m'^s}\\
&=\sum_{n=1}^\infty \sum_{m'=1}^\infty \frac{f(n)g(m'/n)}{m'^s}1_{n\mid m'}\\
&=\sum_{m'=1}^\infty\sum_{n=1}^\infty \frac{f(n)g(m'/n)}{m'^s}1_{n\mid m'}\\
&=\sum_{m'=1}^\infty\sum_{n\mid m'}\frac{f(n)g(m'/n)}{m'^s}\\
&=\sum_{m'=1}^\infty \frac{f*g(m')}{m'^s}.
\end{align}
Here we use the absolutely convergence to interchange the sum and $1_{n\mid m'}$ denotes a function which is $1$ if $n$ divides $m'$ and $0$ otherwise.
