Infinite products

Product related to a series expansion of $\cot(\pi x)$

Today we show the proof of this nice result proposed by @replicamethod \[ \prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \frac{2\pi}{3\sqrt{3}} \] The proof will rely on the expansion in zeta numbers of the $\cot(\pi x)$ function.

Proof

Consider the infinite product \[ A = \prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \prod_{k=1}^{\infty} \frac{k^2}{k^2-\frac{1}{9}} = \prod_{k=1}^{\infty} \frac{1}{1-\frac{1}{9k^2}} = \frac{1}{\prod_{k=1}^{\infty}1-\frac{1}{9k^2}}\] Taking logarithm in both sides: \[ \ln(A) = -\sum_{k=1}^{\infty} \ln\left(1-\frac{1}{9k^2}\right) = \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n9^nk^{2n}} = \underbrace{\sum_{n=1}^{\infty}\frac{1}{n9^n}\sum_{k=1}^{\infty}}_{\textrm{Rearrangement}}\frac{1}{k^{2n}} = \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n9^n}\] Now using the following expansion (proof in the Appendix): \[\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n} x^{2n} = \ln(\pi x) + \ln(\csc \pi x)\] If we put $\displaystyle x = \frac{1}{3}$ \[\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n9^n} = \ln\left(\frac{\pi}{3} \right) + \ln\left(\frac{2}{\sqrt{3}}\right)\] Hence \[ \ln(A) = \ln\left(\frac{\pi}{3} \right) + \ln\left(\frac{2}{\sqrt{3}}\right) = \ln\left(\frac{2\pi}{3\sqrt{3}}\right)\] \[ \Longrightarrow A = \frac{2\pi}{3\sqrt{3}}\] Hence \[ \boxed{ \prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \frac{2\pi}{3\sqrt{3}}} \]

Appendix

Proposition: \[\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k}= \ln\left(\frac{\pi x}{\sin(\pi x)}\right) \quad 0\lt x\lt 1 \] Proof:

Recall the expansion in partial fractions of $\cot(x\pi)$ \[ \cot(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x}+ \sum_{n=1}^{\infty} \frac{x}{(x^2-n^2)}\right] \quad x\neq \pm 1, \pm 2, \pm 3,.. \] From these we can obtain the exapnsion in zeta numbers. Suppose that $|x|\lt1$, using the first equation we have \begin{align*} \cot(x \pi) =& \frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{(n^2-x^2)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2\left(1-\left(\frac{x}{n}\right)^2\right)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2} \sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j}\\ =&\frac{1}{\pi x}- \frac{2}{\pi}\underbrace{\sum_{j=0}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^{2j+2}} x^{2j+1}}_{\textrm{Rearrangement}} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{j=0}^{\infty}\zeta(2j+2) x^{2j+1} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad (j = k-1)\\ \end{align*} Therefore \[\boxed{ \cot(\pi x) =\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad |x|\lt1 } \] Integrating the expansion of $\pi \cot(\pi x)$: \[ \int \pi \cot(\pi x) dx =\int \frac{1}{ x}dx- 2\sum_{k=1}^{\infty}\zeta(2k) \int x^{2k-1}dx + C \] Given that \[ \int \pi \cot(\pi x) dx = \ln(\sin(\pi x))+ C\] Hence \[ \ln(\sin(\pi x)) = \ln(x) -\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + C \] \[ \ln\left(\frac{\sin(\pi x)}{x}\right) = -\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + C\] Taking the limit as $x \to 0$: \[ \ln\left( \lim_{x \to 0 } \frac{\sin(\pi x)}{x}\right) = \underbrace{-\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} }_{=0}+ C \] Given that $\displaystyle \lim_{x \to 0} \frac{\sin(\pi x) }{x} =\pi \Longrightarrow C = \ln(\pi) $ \[ \therefore \ln(\sin(\pi x)) = \ln(x) - \sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + \ln(\pi)\Longrightarrow \ln\left(\frac{\pi x}{\sin(\pi x)}\right) = \sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} \] \[\boxed{\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k}= \ln\left(\frac{\pi x}{\sin(\pi x)}\right) }\]