Are there any values of $m$ other than $3$ such that $\prod_{k=2}^{\infty}\dfrac{(k^m-1)}{ (k^m+1)} $ has some kind of known form?

Question

It has been shown here that $\prod_{k=2}^{\infty}\dfrac{(k^3-1)}{(k^3+1)} =\dfrac23 $.

My question:

Are there any values of $m$ other than $3$ such that $\prod_{k=2}^{\infty}\dfrac{(k^m-1)}{(k^m+1)} $ has some kind of known form?

I can show that $\ln\left(\prod_{k=2}^{\infty}\dfrac{(k^m-1)}{(k^m+1)}\right) =-2\sum_{j=1}^{\infty}\dfrac1{2j-1}(\zeta(m(2j-1))-1) $ but that does not seem to help, though it certainly makes computing the values easier.

Answer 1

The answer depends on the parity of $m$.

For $m$ even, we have $$ \prod _{k=2}^{\infty} \frac{k^m-1}{k^m+1} = \frac{2 \pi i}{m}\prod _{j=1}^{m-1}\left( \sin \left(\pi (-1)^{j/m}\right)\right)^{(-1)^j} $$In particular, for $m=2$ this gives $\pi i \csc(\pi i)=\pi \text{ csch}(\pi) $. When $m$ is odd, we have $$ \prod _{k=2}^{\infty} \frac{k^m-1}{k^m+1} = 2 \prod _{j=1}^{m-1} \frac{\Gamma \left(2+(-1)^j (-1)^{j/m}\right)}{\Gamma \left(2-(-1)^j (-1)^{j/m}\right)} $$In particular, for $m=3$ this gives $2/3$. The proof depends on cancellations, as hinted in Oscar Lanzi's answer; I can expand on this if there's interest.

For instance, this allows for direct calculation of several values: $$ \prod _{k=2}^{\infty} \frac{k^4-1}{k^4+1} =\frac{\pi \sinh (\pi )}{\cosh \left(\sqrt{2} \pi \right)-\cos \left(\sqrt{2} \pi \right)} $$ $$ \prod _{k=2}^{\infty} \frac{k^6-1}{k^6+1} =\frac{2 \pi \cosh ^2\left(\frac{\sqrt{3} \pi }{2}\right) \text{csch}(\pi )}{3 \cosh (\pi )-3 \cos \left(\sqrt{3} \pi \right)} $$

Answer 2

Tl, dr: for $m=2$ the product is $\pi/(\sinh\pi)\approx 0.2720$.

For any integer $m$, we can render a result in terms of the $\Gamma$ function evaluated at the zeroes of the denominator. In the cases $m=2$ and $m=3$, because the involved zeroes are integers or paired up to give integer sums, an elementary form is available. We shall focus on $m=2$, as the value for $m=3$ is already available by simpler means.

For each $k$ we render

$k^2-1=(k+1)(k-1)=(\Gamma(k+2)/\Gamma(k+1))(\Gamma(k)/\Gamma(k-1))$

$k^2+1=(k+i)(k-i)=(\Gamma(k+1+i)/\Gamma(k+i))(\Gamma(k+1-i)/\Gamma(k-i))$

These ratios telescope in the product given in the problem so that the partial product of $n$ terms (up to $k=n+1$) is

$\dfrac{(\Gamma(n+3)/\Gamma(3))(\Gamma(n+1)/\Gamma(1)) }{(\Gamma(n+2+i)/\Gamma(2+i))(\Gamma(n+2-i)/\Gamma(2-i))}$

$=\dfrac{\Gamma(n+3)\Gamma(n+1)\Gamma(2+i)\Gamma(2-i)}{2\Gamma(n+2+i)\Gamma(n+2-i)}$

Now for the sneaky part. From the Stirling approximation for the $\Gamma$ function which applies for all paths to infinity with limiting argument $0\le\theta<\pi$, we infer that

$\Gamma(n+\alpha)\sim n^\alpha\Gamma(n),n\to\infty$.

This causes $\Gamma(n+3)\Gamma(n+1)$ to cancel $\Gamma(n+2+i)\Gamma(n+2-i)$ in our fraction leading to the complete product

$\dfrac{\Gamma(2+i)\Gamma(2-i)}{2}$

Now, we have to render the product in the numerator as elementary functions. This is where the pairing of zeroes into integer sums comes in; because the zeroes of $k^2+1$ add up to an integer so do the arguments of the remaining $\Gamma$ functions. Which then allows us to apply the reflection formula. Render

$\Gamma(2+i)\Gamma(2-i)=(1+i)(i)(-1+i)\color{blue}{\Gamma(-1+i)\Gamma(2-i)}$

where the arguments in the blue terms now add up to $1$ and the reflection formula gives

$\Gamma(2+i)\Gamma(2-i)=(1+i)(i)(-1+i)\color{blue}{\pi/\sin(-\pi+i\pi)}=(-2i\pi)/(-i\sinh\pi)=2\pi/(\sinh\pi)$

Our product is half of that, thus $\color{blue}{\pi/(\sinh\pi)\approx 0.2720}$.