Euler Product
Theorem
Let $\displaystyle f(s) = \sum_{n \in \N} a_n n^{-s}$ be a Dirichlet series, absolutely convergent on $\Re(s) > \sigma_a$ (see Abscissa of Absolute Convergence).
- $\displaystyle \sum_{n=1}^\infty a_n n^{-s} = \prod_p \frac 1 {1-a_p p^{-s}} $
for all $s$ with $\Re(s) > \sigma_a$, where $p$ ranges over the primes.
This representation for $f$ is called an Euler product for the Dirichlet series.
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Completely multiplicative hypothesis not mentioned. Needs also the statement: $\displaystyle f(s) = \prod_p\left\{ \sum_{k \geq 1} a_{p^k}p^{-ks}\right\}$ or however it goes for multiplicative but not completely so functions...(know this is messy, just a note-to-self; I'll fix it soon)
Proof
This is immediate from Product Form of Sum on Completely Multiplicative Function.
Source of Name
This entry was named for Leonhard Paul Euler.
