Dirichlet series of Dirichlet product equals product of Dirichlet series

Statement

For formal Dirichlet series

Suppose $f$ and $g$ are arithmetic functions. Then, the Dirichlet series of the Dirichlet product $f*g$ equals the product of the Dirichlet series for $f$ and for $g$.

In symbols:

$\sum _{n\in \mathbb {N} }{\frac {f(n)}{n^{s}}}\sum _{n\in \mathbb {N} }{\frac {g(n)}{n^{s}}}=\sum _{n\in \mathbb {N} }{\frac {(f*g)(n)}{n^{s}}}$.

For functions

This identity holds for the functions corresponding to the Dirichlet series whenever these series are absolutely convergent, because the additive rearrangement needed is valid when the series are absolutely convergent. Note that this also shows that if the Dirichlet series for $f$ and $g$ are absolutely convergent for some $s$, the Dirichlet series for $f*g$ is also absolutely convergent for the same $s$.

Finally, if we consider the analytic continuations of these Dirichlet series to the complex numbers, then the analytic continuation for the function corresponding to $f*g$ is the product of the analytic continuations for $f$ and for $g$.

Related facts

• Product formula for Dirichlet series of completely multiplicative function
• Product formula for Dirichlet series of multiplicative function

Proof

We expand the right side:

$\sum _{n\in \mathbb {N} }{\frac {(f*g)(n)}{n^{s}}}=\sum _{n\in \mathbb {N} }\sum _{d_{1}d_{1}=n}{\frac {f(d_{1})}{d_{1}^{s}}}{\frac {g(d_{2})}{d_{2}^{s}}}$.

Note that every ordered pair $(d_{1},d_{2})$ occurs for exactly one $n$, namely $n=d_{1}d_{2}$. Thus, the above sum can be rearranged as:

$\sum _{n\in \mathbb {N} }{\frac {(f*g)(n)}{n^{s}}}=\sum _{d_{1}\in \mathbb {N} }{\frac {f(d_{1})}{d_{1}^{s}}}\sum _{d_{2}\in \mathbb {N} }{\frac {g(d_{2})}{d_{2}^{s}}}$.

Relabeling $d_{1}$ and $d_{2}$, we get the identity we need to prove.