An infinite product obtained by taking every third term from the Wallis product. Is it new?

Question

Numerical calculation suggests that $\prod_{n=0}^\infty \frac{(6n+2)^2}{(6n+1)(6n+3)} = \frac{3^{1/2}}{2^{1/3}}$.

Has this been proven?

Answer 1

Let $ n $ be a positive integer, we have : \begin{aligned} \frac{\Gamma^{2}\left(k+1+\frac{1}{3}\right)}{\Gamma\left(k+1+\frac{1}{6}\right)\Gamma\left(k+1+\frac{1}{2}\right)}&=\frac{\left(6k+2\right)^{2}}{\left(6k+1\right)\left(6k+3\right)}\times\frac{\Gamma^{2}\left(k+\frac{1}{3}\right)}{\Gamma\left(k+\frac{1}{6}\right)\Gamma\left(k+\frac{1}{2}\right)}\\ \Longrightarrow\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\Gamma^{2}\left(n+\frac{4}{3}\right)}{\Gamma\left(n+\frac{7}{6}\right)\Gamma\left(n+\frac{3}{2}\right)}&=\prod_{k=0}^{n}{\frac{\left(6k+2\right)^{2}}{\left(6k+1\right)\left(6k+3\right)}}\times\frac{\Gamma^{2}\left(\frac{1}{3}\right)}{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{2}\right)} \end{aligned}

Using the Legendre duplication formula, we have that : \begin{aligned} \Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{3}+\frac{1}{2}\right)&=\sqrt[3]{2}\sqrt{\pi}\;\Gamma\left(\frac{2}{3}\right)\\ \iff \Gamma\left(\frac{1}{3}\right)\Gamma\left(1-\frac{1}{6}\right)&=\sqrt[3]{2}\sqrt{\pi}\;\Gamma\left(1-\frac{1}{3}\right)\end{aligned}

Multiplying both sides by $ \Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right) $, then using Euler's reflection formula, we get : \begin{aligned} \Gamma^{2}\left(\frac{1}{3}\right)\times\frac{\pi}{\sin{\left(\frac{\pi}{6}\right)}}&=\sqrt[3]{2}\sqrt{\pi}\;\Gamma\left(\frac{1}{6}\right)\times\frac{\pi}{\sin{\left(\frac{\pi}{3}\right)}}\\ \iff \Gamma^{2}\left(\frac{1}{3}\right)\times 2\pi&=\sqrt[3]{2}\;\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)\times\frac{2\pi}{\sqrt{3}}\\ \iff \ \ \frac{\Gamma^{2}\left(\frac{1}{3}\right)}{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{2}\right)}&=\frac{\sqrt[3]{2}}{\sqrt{3}} \end{aligned}

Thus, for every positive integer $ n $ we have : $$ \prod_{k=0}^{n}{\frac{\left(6k+2\right)^{2}}{\left(6k+1\right)\left(6k+3\right)}}=\frac{\sqrt{3}\;\Gamma^{2}\left(n+\frac{4}{3}\right)}{\sqrt[3]{2}\;\Gamma\left(n+\frac{7}{6}\right)\Gamma\left(n+\frac{3}{2}\right)} $$

Using Stirling's formula, we get the Following approximations : \begin{aligned}\Gamma\left(n+\frac{4}{3}\right)&\underset{n\to +\infty}{\sim}\sqrt{2\pi\left(n+\frac{1}{3}\right)}\left(\frac{n+\frac{1}{3}}{\mathrm{e}}\right)^{n+\frac{1}{3}}\\ \Gamma\left(n+\frac{7}{6}\right)&\underset{n\to +\infty}{\sim}\sqrt{2\pi\left(n+\frac{1}{6}\right)}\left(\frac{n+\frac{1}{6}}{\mathrm{e}}\right)^{n+\frac{1}{6}}\\ \Gamma\left(n+\frac{3}{2}\right)&\underset{n\to +\infty}{\sim}\sqrt{2\pi\left(n+\frac{1}{2}\right)}\left(\frac{n+\frac{1}{2}}{\mathrm{e}}\right)^{n+\frac{1}{2}}\end{aligned}

Which lead to the following : $$ \frac{\Gamma^{2}\left(n+\frac{4}{3}\right)}{\Gamma\left(n+\frac{7}{6}\right)\Gamma\left(n+\frac{3}{2}\right)}\underset{n\to +\infty}{\sim}1 $$

Thus : $$ \prod_{n=0}^{+\infty}{\frac{\left(6n+2\right)^{2}}{\left(6n+1\right)\left(6n+3\right)}}=\lim_{n\to +\infty}{\prod_{k=0}^{n}{\frac{\left(6k+2\right)^{2}}{\left(6k+1\right)\left(6k+3\right)}}}=\frac{\sqrt{3}}{\sqrt[3]{2}} $$