Sum of Reciprocals of Powers as Euler Product/Corollary 2
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Corollary to Sum of Reciprocals of Powers as Euler Product
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\sigma > 1$.
Then:
- $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-s} } {1 - p^{-s} } } = \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }$
where the infinite product runs over the prime numbers.
Proof
\(\ds \prod_{\text {$p$ prime} } \frac 1 {\paren {1 - p^{-s} }^2}\) | \(=\) | \(\ds \paren {\map \zeta s}^2\) | Sum of Reciprocals of Powers as Euler Product | |||||||||||
\(\ds \prod_{\text {$p$ prime} } \frac 1 {1 - p^{-2 s} }\) | \(=\) | \(\ds \map \zeta {2 s}\) | ||||||||||||
\(\ds \prod_{\text {$p$ prime} } \frac {\paren {1 - p^{-2 s} } } {\paren {1 - p^{-s} }^2}\) | \(=\) | \(\ds \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }\) | ||||||||||||
\(\ds \prod_{\text {$p$ prime} } \frac {\paren {1 + p^{-s} } \paren {1 - p^{-s} } } {\paren {1 - p^{-s} }^2}\) | \(=\) | \(\ds \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }\) | ||||||||||||
\(\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-s} } {1 - p^{-s} } }\) | \(=\) | \(\ds \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }\) |
$\blacksquare$
Examples
Example: $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-2} } {1 - p^{-2} } }$
- $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-2} } {1 - p^{-2} } } = \dfrac 5 2$
Example: $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-4} } {1 - p^{-4} } }$
- $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-4} } {1 - p^{-4} } } = \dfrac 7 6$