Euler Product
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Completely multiplicative hypothesis not mentioned. Needs also the statement: $\displaystyle f \left({s}\right) = \prod_p \left\{{\sum_{k \mathop \ge 1} a_{p^k} p^{-k s}}\right\}$ or however it goes for multiplicative functions which are not completely multiplicative
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Theorem
Let $a_n : \N \to \C$ be an arithmetic function.
Let $\displaystyle f \left({s}\right) = \sum_{n \mathop \in \N} a_n n^{-s}$ be its Dirichlet series.
Let $\sigma_a$ be its abscissa of absolute convergence.
Then for $\Re \left({s}\right) > \sigma_a$:
- $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-s} = \prod_p \frac 1 {1 - a_p p^{-s} }$
where $p$ ranges over the primes.
This representation for $f$ is called an Euler product for the Dirichlet series.
Completely multiplicative hypothesis not mentioned. Needs also the statement: $\displaystyle f \left({s}\right) = \prod_p \left\{{\sum_{k \mathop \ge 1} a_{p^k} p^{-k s}}\right\}$ or however it goes for multiplicative functions which are not completely multiplicative
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Proof
This is immediate from Product Form of Sum on Completely Multiplicative Function.
Source of Name
This entry was named for Leonhard Paul Euler.