User talk:Poliagapitos

Infinite product
Hi, I would like to ask if you could add a reference or proof for the result $$\frac{\sqrt{3}\pi}{6}=\left(\displaystyle\prod_{p \equiv 1 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p-1}\right) \cdot \left(\displaystyle\prod_{p \equiv 5 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p+1}\right)=\frac{5}{6} \cdot \frac{7}{6} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{18} \cdots ,$$ which you added on the wikipedia article List of formulae involving π in a edited made on June, thanks in advance and best regards Dabed (talk) 03:53, 26 October 2021 (UTC)