Sum of Reciprocals of Powers as Euler Product/Proof 1

Theorem

Let $\zeta$ be the Riemann zeta function.

Let $s \in \C$ be a complex number with real part $\sigma > 1$.

Then:

$\ds \map \zeta s = \prod_{\text {$p$ prime} } \frac 1 {1 - p^{-s} }$

where the infinite product runs over the prime numbers.

Proof

By definition of Euler product:

$\ds \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$

if and only if $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is absolutely convergent.

This article, or a section of it, needs explaining. In particular: I believe only "if" is true as shown in Product Form of Sum on Completely Multiplicative Function. Or is there a proof for "only if"? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

For all $n \in \Z_{\ge 1}$, let $a_n = 1$.

From Convergence of P-Series:

$\ds \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent

if and only if:

$\cmod z > 1$

It then follows that:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$

$\blacksquare$