First, we need there is a key fact you may or may not have seen in calculus. Roughly speaking, when series converge absolutely, you can mess around with them with a lot with impunity. (See, for instance, Mertens' Theorem on convergence of Cauchy products; even [C.3.6] doesn't say any more than this in discussing this proof.) Since \(F\) and \(G\) converge absolutely, we will mess around a lot with the product \begin{equation*}F(s)G(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}\sum_{m=1}^\infty\frac{g(m)}{m^s}\end{equation*} In particular, we can group the products by the terms \(\frac{f(n)g(m)}{n^sm^s}\) (the same way we did in proving things about \(\star\) in Subsection 23.4.3), without loss of equality.

We can further group by when \(n\) and \(m\) are complementary divisors of the same number. This gives \begin{equation*}F(s)G(s)=\sum_{d=1}^\infty\sum_{nm=d}\frac{f(m)g(n)}{d^s}\; .\end{equation*}

Notice that the inner sum is precisely the Dirichlet \(\star\) product (except divided by \(d^s\)). So we may rewrite this as \begin{equation*}F(s)G(s)=\sum_{d=1}^\infty \frac{(f\star g)(d)}{d^s}\; .\end{equation*}

The numerators are the definition of \(h\), so this is just \(H(s)\), as desired.