Closed form for $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n{\left(2-\frac{2n^2-\pi^2+8}{n^2}\cos{\frac{(2k-1)\pi}{n}}\right)}$?

Question

I am looking for a closed form for: $$\lim_{n\to\infty}\prod_{k=1}^n{\left(2-\frac{2n^2-\pi^2+8}{n^2}\cos{\frac{(2k-1)\pi}{n}}\right)}$$

(Wolfram suggests that it's approximately 6.17966.)

Context:
I was thinking about the curve $y=2^{n-1}(x-\cos{\frac{0\pi}{n}})(x-\cos{\frac{1\pi}{n}})(x-\cos{\frac{2\pi}{n}})...(x-\cos{\frac{n\pi}{n}})$.

Here is the graph when $n=6$, for example:

It is an $n+1$ degree curve tangent to the unit circle at $n$ points, and the total area of the $n$ regions enclosed by the curve and the x-axis is always $1$ (for any $n$).

The product, let's call it $P$, of the areas of the regions enclosed by the curve and the x-axis, approaches $0$ as $n\to\infty$. But then I discovered that if we magnify the curve so that the average area of the regions is always $2$ (for any $n$), then $P$ approaches 6.17966... as $n\to\infty$. The limit in the question is this number. (To derive the limit expression, I used an equivalent expression for the curve: $y=-\sqrt{1-x^2}\sin{(n\arccos{x})}$, and stretched it vertically and horizontally by scale factor $\sqrt{2n}$ so that the average area of the regions is $2$.)

This idea of the product of areas of regions in a circle approaching some positive number is inspired by this question.

My attempt:
I tried relating the limit to a similar limit, $\lim_{n\to\infty}\prod_{k=1}^n{\left(2-2\cos{\frac{(2k-1)\pi}{n}}\right)}=4$, as well as taking the log of the product, but I didn't make any progress. I am not very familiar with evaluating infinite products and would like to learn more.

Answer 1

The closed form of the product is basically $$\prod_{k=1}^n\left(1-2z\cos\frac{(2k-1)\pi}{n}+z^2\right)=(1+z^n)^2,$$ shown using $(z-e^{i\phi})(z-e^{-i\phi})=1-2z\cos\phi+z^2$ and the roots of $z^n+1=0$.

Thus, to compute $f(\alpha)=\displaystyle\lim_{n\to\infty}\prod_{k=1}^n\big[2-(2-\alpha^2/n^2)\cos\big((2k-1)\pi/n\big)\big]$, we choose $z=z_n$ so that $1-2z\cos\phi+z^2$ is proportional to $2-(2-\alpha^2/n^2)\cos\phi$: $$\frac{2z_n}{1+z_n^2}=r_n:=1-\frac{\alpha^2}{2n^2}\implies z_n=\frac{1\pm\sqrt{1-r_n^2}}{r_n}.$$ Now we easily get $z_n=1\pm\alpha/n+o(1/n)$ as $n\to\infty$, hence $$f(\alpha)=\lim_{n\to\infty}\left(\frac2{1+z_n^2}\right)^n(1+z_n^n)^2=e^\alpha(1+e^{-\alpha})^2.$$

Your limit is $f\left(\sqrt{\pi^2-8}\right)\approx6.17966119933834352235124640584393+$.