Product of two absolutely convergent Dirichlet series

Question

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:$$F(s)G(s) = \left( \sum_{n=1}^\infty f(n)n^{-s}\right)\left( \sum_{n=1}^\infty g(n)n^{-s}\right) = \sum_{n=1}^\infty (f * g)(n)n^{-s}?$$

Answer 1

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.

Answer 2

$$ \sum_{n\geq 1}\frac{(f*g)(n)}{n^s} = \sum_{n\geq 1}\frac{1}{n^s}\sum_{d\mid n}f(d)g(n/d) = \sum_{k\geq 1} \sum_{d\geq 1}\frac1{(kd)^s}f(d)g(k) \;. $$