An Infinite product

[Shobhit]

Show that
\[\prod_{k=2}^{\infty} \left(\frac{2k+1}{2k-1}\right)^{k} \left(1-\frac{1}{k^2}\right)^{k^2}=\frac{\sqrt{2}}{6}\pi \]

This problem can be solved using only elementary methods.

 

[Shobhit]

Spoiler

Here are some hints regarding this problem:

Start by showing that

\(\displaystyle \prod_{k=2}^{N} \left(\frac{2k+1}{2k-1}\right)^{k}\left(1-\frac{1}{k^2}\right)^{k^2}= \displaystyle \frac{1}{6}\frac{(N!)^3}{(2N)!} \frac{2^N (2N+1)^N (N+1)^{N^2}}{N^{N^2+2N+1}} \)

Then let $N\to \infty$ and use Stirling's approximation to evaluate the limit.

\begin{align*} \prod_{k=2}^{\infty} \left(\frac{2k+1}{2k-1}\right)^{k} \left(1-\frac{1}{k^2}\right)^{k^2} &= \frac{1}{6}\lim_{N\to \infty} \frac{(N!)^3}{(2N)!} \frac{2^N (2N+1)^N (N+1)^{N^2}}{N^{N^2+2N+1}} \\ &= \frac{\sqrt{2} \pi}{6}\lim_{N\to \infty}\frac{(N+1)^{N^2}(2N+1)^N}{e^{N} N^{N^2+N} 2^N} \\ &= \frac{\sqrt{2}}{6}\pi \end{align*}