Let's consider the following function: $$ f(n, k) = \prod_{i = 2}^{n}{\frac{i^k - 1}{i^k + 1}} $$ I'm interested in the limit of $f$ as $n$ approaches infinity for various values of $k$.
I managed to calculate the limit for $k = 1$ (it's $2$) and $k = 3$ (it's $2/3$).
For other values of $k$ the limit seems to be irrational, but I only managed to approximate it with direct calculations.
My question:
How to find the limit analytically for other $k$? What's so special about $3$ and $1$ being rational limits?
When $k=1$,
$$ \lim_{n\to\infty} f(n,1) = \lim_{n\to\infty} \frac{2}{n(n+1)} = 0. $$
When $k=2,3,\dots$, you may write
$$f(n,k)=\Biggl[\prod_{\omega\,:\,\omega^k=1}\frac{(n-\omega)!}{(1-\omega)!}\Biggr]\Bigg/\Biggl[\prod_{\omega\,:\,\omega^k=-1}\frac{(n-\omega)!}{(1-\omega)!}\Bigg],$$
where $s!=\Gamma(s+1)$. Then the Stirling's approximation shows that the limit as $n\to\infty$ is
$$\lim_{n\to\infty}f(n,k)=\frac{\prod_{j=1}^{k}(1-e^{(2j-1)i\pi/k})!}{\prod_{j=1}^{k}(1-e^{2ji\pi/k})!}.$$
For even values of $k$ as well as $k=3$, this reduces to an elementary expression:
\begin{align*} \lim_{n\to\infty}f(n,2) &= \frac{\pi}{\sinh\pi}, \\ \lim_{n\to\infty}f(n,3) &= \frac{2}{3}, \\ \lim_{n\to\infty}f(n,4) &= \frac{\pi\sinh\pi}{\cosh(\sqrt{2}\pi) - \cos(\sqrt{2}\pi)}, \\ \lim_{n\to\infty}f(n,6) &= \frac{\pi(\cosh(\sqrt{3}\pi)+1)}{3\sinh \pi (\cosh \pi - \cos(\sqrt{3}\pi) )}. \end{align*}