The Euler Product of a Dirichlet Series

# The Euler Product of a Dirichlet Series

 Definition: Let $a : \mathbb{N} \to \mathbb{C}$ be a multiplicative arithmetic function. Then the corresponding Dirichlet series can be written as $\displaystyle{\sum_{n=1}^{\infty} \frac{a(n)}{n^s} = \prod_{p \: \mathrm{prime}} \left ( 1 + \frac{a(p)}{p^s} + \frac{a(p^2)}{p^{2s}} + ... \right )}$.

Note that if $a(n)$ is a completely multiplicative arithmetic function then:

(1)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{a(n)}{n^s} &= \prod_{p \: \mathrm{prime}} \left ( 1 + \frac{a(p)}{p^s} + \frac{a(p^2)}{p^{2s}} + ... \right ) \\ &=\prod_{p \: \mathrm{prime}} \left ( 1 + \frac{a(p)}{p^s} + \frac{a(p)^2}{p^{2s}} + ... \right ) \\ &= \prod_{p \: \mathrm{prime}} \underbrace{\sum_{k=0}^{\infty} \left ( \frac{a(p)}{p^s} \right )^k}_{\mathrm{geometric \: series}} \\ &= \prod_{p \: \mathrm{prime}} \frac{1}{1 - \frac{a(p)}{p^s}} \end{align}
 Proposition 1: $\displaystyle{\zeta (s) = \prod_{p \: \mathrm{prime}} \frac{1}{1 - \frac{1}{p}}}$.
• Proof: The arithmetic function $1(n) = 1$ is completely multiplicative. So it's Euler product has the form:
(2)
\begin{align} \quad \zeta (s) &= \sum_{n=1}^{\infty} \frac{1}{n^s} \\ &= \prod_{p \: \mathrm{prime}} \frac{1}{1 - \frac{1}{p}} \quad \blacksquare \end{align}