Find $C$ such that $\frac{1}{n}\prod_{k=1}^{n}C\left(\cos{\frac{k\pi}{2(n+1)}}+\sin{\frac{k\pi}{2(n+1)}}-1\right)$ converges to a positive number.

Question

I'm looking for the value of $C$ such that $L=\lim\limits_{n\to\infty}\frac{1}{n}\prod\limits_{k=1}^{n}C\left(\cos{\frac{k\pi}{2(n+1)}}+\sin{\frac{k\pi}{2(n+1)}}-1\right)$ equals a positive real number.

Desmos suggests $C\approx 4.5395$. I'm looking for a closed form.

(I'm not so interested in the value of $L$, but I wouldn't mind knowing that also; desmos suggests $L\approx 0.8817$.)

Context: This question is related to another question about an infinite product involving a quarter-circle inscribed in a square. ($C$ in this question seems to equal $\frac{\pi}{4}A$ in the other question.) I think this question is interesting by itself, so I'm asking it here.

My attempt: I have tried to take the log of the product and relate the resulting sum to an integral, but I do not know how to deal with the $\frac{1}{n}$ and also the $(n+1)$. I have considered using complex numbers, but I do not know how that could be done.

Answer 1

One might use a Riemann sum argument, leading to a known integral: $$\log C=-\int_0^1\log\left(\cos\frac{x\pi}2+\sin\frac{x\pi}2-1\right)\,dx.$$

Alternatively, use $C=\lim\limits_{n\to\infty}P_n^{-1/n}$ with $$P_n:=\prod_{k=1}^{n-1}\left(\cos\frac{k\pi}{2n}+\sin\frac{k\pi}{2n}-1\right)=2^{3(n-1)/2}\prod_{k=1}^{n-1}\sin^2\frac{k\pi}{4n}$$ to get an integral reducing to $\int_0^{\pi/4}\log\sin t\,dt$. Thus, using Catalan's constant $G$, $$C=e^{4G/\pi}\sqrt2\approx4.5395015097179854508819422150513221738\cdots$$

Similarly, via Euler–Maclaurin summation formula, one gets $$L=\lim_{n\to\infty}(C^n P_{n+1}/n)=4/C.$$

Answer 2

This is not an answer, but rather an observation that is enabled by @metamorphy's answer.

In a $90^{\text{o}}$ circular sector, draw a line segment connecting the opposite vertices, and draw line segments connecting the central vertex with points uniformly spaced along the arc, thus dividing the sector into $2n$ regions. Here is an example with $n=6$.

If the area of the sector is $\dfrac{e^{4G/\pi}}{\sqrt2}n\approx 2.26975n$, where $G$ is Catalan's constant, then the product of the areas of the regions approaches $2$ as $n\to\infty$.