TITLE: How to find values of $f(m,k)=\prod\limits_{n=1}^{\infty}\left[\frac{(mn-k)^2}{(mn-k+1)(mn-k-1)}\right]$?
QUESTION [8 upvotes]: $f(2,0)=\frac{\pi}{2}$ is very popular result, but there are a lot of them, not involving $\pi$, for example
$$f(4,2)=\prod\limits_{n=1}^{\infty}\left[\frac{(4n-2)^2}{(4n-1)(4n-3)}\right]=\sqrt{2}$$
$$f(6,3)=\prod\limits_{n=1}^{\infty}\left[\frac{(6n-3)^2}{(6n-2)(6n-4)}\right]=\frac{2}{\sqrt{3}}$$
$$f(6,4)=\prod\limits_{n=1}^{\infty}\left[\frac{(6n-4)^2}{(6n-3)(6n-5)}\right]=\frac{3^{1/2}}{2^{1/3}}$$
How to find values of $f(m,k)$ for all $m>1$ and $(m-1)>k\geqslant0$?

REPLY [8 votes]: Hint. One may recall that
$$
\prod_{n=1}^{N} \left(1 + \frac{z}{n+a}\right)=\frac{\Gamma(a+1)\Gamma(a+N+z+1)}{\Gamma(a+N+1)\Gamma(a+z+1)} \tag1
$$ where $\Gamma$ is the Euler gamma function, one may then write
$$
\prod\limits_{n=1}^{N}\left[\frac{(mn-k)^2}{(mn-k+1)(mn-k-1)}\right]=\left[\prod\limits_{n=1}^{N}\left(1 + \frac{1/m}{n-k/m}\right)\left(1 - \frac{1/m}{n-k/m}\right)\right]^{-1} \tag2
$$
and apply $(1)$ to $(2)$, then letting $N \to \infty$.
Edit. The preceding steps give

$$
\prod\limits_{n=1}^{\infty}\left[\frac{(mn-k)^2}{(mn-k+1)(mn-k-1)}\right]=\frac{\Gamma \left(1-\frac{k+1}{m}\right) \Gamma \left(1-\frac{k-1}{m}\right)}{\Gamma \left(1-\frac{k}{m}\right)^2}
$$

from which one deduces many particular cases.