Let $f$ and $g$ be an arithmetic functions, and let $f*g$ be the Dirichlet convolution of $f$ and $g$.
As known from fundamental analytic number theory, the Dirichlet series generating function is: $DG(f;s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$.
Hence, The multiplication of Dirichlet series can be written as: $DG(f;s)DG(g;s)=DG(f*g;s)$, but as we know, Riemann series theorem says that if the series is conditionally convergence, then any permutation of the series may generate another sum.
My question is, if $DG(f;s)$ $DG(g;s)$ are conditionally convergence, how can we definde $DG(f;s)DG(g;s)$?
This question can be completely answered by the following surprising theorem.
Theorem: Let $D(f,s)$, $D(g,s)$ be the Dirichlet series corresponding to $f$ and $g$. If $D(f,s)$ and $D(g,s)$ conditionally converge at $s=0$, then the convolution Dirichlet series $D(f*g,s)$ must be conditionally converges at $s=\frac{1}{2}+it$ for each $t\in \mathbb{R}$. And the constant "$\frac{1}{2}$" is optimal as concerns $\text{Re}\hspace{0.05cm}s$.
The proof can be found in the book Diophtine apprximation and Dirichlet series,pp114 theorem 4.3.4 for second Edition.
I really like its proof, which used hyperbola method of Dirichlet to prove the convergence, and used Banach-Steinhuas theorem and Kronecker's lemma to prove the optimality.
Other interesting consequences can also be found in section 4.3 of this book.