Product of Dirichlet Series

Theorem

Let \(f\) and \(g\) be arithmetic functions and consider the Dirichlet series'

\[ \sum_{n = 1}^\infty \frac{f(n)}{n^s} \quad \text{and} \quad \sum_{n = 1}^\infty \frac{g(n)}{n^s}.\]

Formally, their product is given by

\[ \left(\sum_{n = 1}^\infty \frac{f(n)}{n^s}\right)\left(\sum_{n = 1}^\infty \frac{g(n)}{n^s}\right) = \sum_{k = 1}^\infty \frac{(f \ast g)(k)}{k^s}\]

and this product converges absolutely if both series converge absolutely.

Proof

Formally,

\[\begin{align*} \left(\sum_{n = 1}^\infty \frac{f(n)}{n^s}\right)\left(\sum_{n = 1}^\infty \frac{g(n)}{n^s}\right) &= \sum_{(m, n) \in \mathbb{Z}_{> 0}^2} \frac{f(n)g(m)}{(nm)^s} \\ &= \sum_{k = 1}^\infty \sum_{\substack{(m, n) \in \mathbb{Z}_{> 0}^2 \\ mn = k }} \frac{f(n)g(m)}{(nm)^s} \\ &= \sum_{k = 1}^\infty \frac{1}{k^s} \sum_{\substack{(m, n) \in \mathbb{Z}_{> 0}^2 \\ mn = k }} f(n)g(m) \\ &= \sum_{k = 1}^\infty \frac{1}{k^s} \sum_{n \mid k} f(n)g\left(\frac{k}{n}\right) \\ &= \sum_{k = 1}^\infty \frac{(f \ast g)(k)}{k^s}. \\ \end{align*}\]

If both series converge absolutely, then the convergence of the formal series follows from this result.