Sum of Reciprocals of Powers as Euler Product/Corollary 2

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$