Zeta of 2 as Product of Fractions with Prime Numerators

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Theorem

\(\ds \map \zeta 2\) \(=\) \(\ds \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac 2 1 \times \dfrac 2 3 \times \dfrac 3 2 \times \dfrac 3 4 \times \dfrac 5 4 \times \dfrac 5 6 \times \dfrac 7 6 \times \dfrac 7 8 \times \dfrac {11} {10} \times \dfrac {11} {12} \times \dfrac {13} {12} \times \dfrac {13} {14} \times \cdots\)

where:

$\zeta$ denotes the Riemann zeta function
$\ds \prod_p$ denotes the continued product over all prime numbers.


Proof

From Sum of Reciprocals of Powers as Euler Product:

$\ds \map \zeta z = \prod_p \frac 1 {1 - p^{-z} }$

where $p$ ranges over the prime numbers.


Thus:

\(\ds \map \zeta 2\) \(=\) \(\ds \prod_p \frac 1 {1 - p^{-2} }\)
\(\ds \) \(=\) \(\ds \prod_p \frac {p^2} {p^2 - 1}\) multiplying top and bottom by $p^2$
\(\ds \) \(=\) \(\ds \prod_p \frac {p^2} {\paren {p - 1} \paren {p + 1} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }\)

which is the result required.

$\blacksquare$


Sources

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