What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?

Question

What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?

As far as I can tell, this has no closed-form solution (not saying much, I don't know much math), but a friend of mine swears he saw a closed-form solution to this in some text he doesn't remember.

Running it through WolframAlpha gives me an approximation which, entered through an inverse symbolic calculator, gets no results. This is satisfactory enough for me to believe there is no closed-form solution, but my friend really does insist there is one.

Edit

Okay, not meaning to sound greedy, but I'd also be interested in seeing how I'd derive a closed-form solution algebraically. I do appreciate the fast answers, though.

Answer 1

Start with the infinite product expansion of $\sin x$,

$$\sin x = x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2\pi^2}\right)$$

We get

$$\prod_{n=2}^\infty \left(1 - \frac{\alpha^2}{n^2}\right) = \frac{\sin(\alpha\pi)}{\alpha\pi(1-\alpha^2)}$$

In particular, $$\begin{align} \prod_{n=2}^\infty \left(1 - \frac{1}{n^2}\right) &= \lim_{\alpha\to 1}\frac{\sin(\alpha\pi)}{\alpha\pi(1-\alpha^2)} = \lim_{a\to1} \frac{\pi\cos(\pi\alpha)}{\pi\alpha(-2\alpha)} = \frac{1}{2}\\ \prod_{n=2}^\infty \left(1 + \frac{1}{n^2}\right) &=\frac{\sin(i\pi)}{i\pi(1-i^2)} = \frac{\sinh\pi}{2\pi}\\ \prod_{n=2}^\infty \left(1 \mp \frac{i}{n^2}\right) &= \frac{\sin(\frac{1\pm i}{\sqrt{2}}\pi)}{\frac{1\pm i}{\sqrt{2}}\pi(1 \mp i)} = \frac{ \sin(\frac{\pi}{\sqrt{2}})\cosh(\frac{\pi}{\sqrt{2}}) \pm i \cos(\frac{\pi}{\sqrt{2}})\sinh(\frac{\pi}{\sqrt{2}}) }{\frac{1\pm i}{\sqrt{2}}\pi(1 \mp i)} \end{align}$$

From this, we get

$$\begin{align} \prod_{n=2}^\infty\frac{n^4-1}{n^4+1} &= \frac{\displaystyle \prod_{n=2}^\infty \left(1 - \frac{1}{n^2}\right) \prod_{n=2}^\infty \left(1 + \frac{1}{n^2}\right) }{\displaystyle \prod_{n=2}^\infty \left(1 - \frac{i}{n^2}\right) \prod_{n=2}^\infty \left(1 + \frac{i}{n^2}\right) }\\ &= \frac{\displaystyle \frac{\sinh\pi}{4\pi} }{\displaystyle\left( \frac{\sin(\frac{\pi}{\sqrt{2}})^2\cosh(\frac{\pi}{\sqrt{2}})^2 +\cos(\frac{\pi}{\sqrt{2}})^2\sinh(\frac{\pi}{\sqrt{2}})^2}{2\pi^2}\right) }\\ &= \frac{\pi\sinh\pi}{\cosh(\sqrt{2}\pi) - \cos(\sqrt{2}\pi)}\\ \\ &\approx 0.84805404935290039212965018340500770584798748\ldots \end{align}$$

Answer 2

$$\prod_{n\geq2}\frac{n^4-1}{n^4+1} = \frac{\pi\sinh(\pi)}{\cosh(\sqrt2\pi)-\cos(\sqrt2\pi)}$$

Answer 3

Here is a closed form

$$P = {\frac {\pi\sinh \left( \pi \right) }{2\sin ( \pi\alpha )) \sin \left(\pi \overline \alpha)\right) }}\approx .8480540500, $$

where $\alpha = \pi \,\left( \frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2} \right) .$

Answer 4

Since your post was interesting and the answers really nice, just for personal curiosity, I looked at the more general function $$\prod_{n=2}^\infty\frac{n^q-1}{n^q+1} $$ where $q$ in an integer. I report below some results I found interesting $$\prod_{n=2}^\infty\frac{n^2-1}{n^2+1}=\pi \text{csch}(\pi )$$ $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}=\frac{2}{3}$$ $$\prod_{n=2}^\infty\frac{n^4-1}{n^4+1}=\frac{\pi \sinh (\pi )}{\cosh \left(\sqrt{2} \pi \right)-\cos \left(\sqrt{2} \pi \right)}$$ $$\prod_{n=2}^\infty\frac{n^6-1}{n^6+1}=\frac{\pi \left(1+\cosh \left(\sqrt{3} \pi \right)\right) \text{csch}(\pi )}{3 \left(\cosh (\pi )-\cos \left(\sqrt{3} \pi \right)\right)}$$