Interesting combo.

For $\frac 1{\omega_n} =\prod_{k=1}^{n}\cot^{2}\left(\dfrac{k\pi}{2n+1}\right)\left(\sum_{k=0}^{2n}\cot\left(x+\frac{k\pi}{2n+1}\right)\right)$

Find$\Omega =\lim_{n\to\infty}\left(\lim_{x\to\frac{\pi}{4n+2}}\left(\dfrac{\tan x}{\tan\left(\frac{\pi}{4n+2}\right)}\right)^{\omega_n}\right)$ This is the proposal by teachers Marian Ursărescu and Florică Anastase, Romania and published by Romanian Mathematical Magazine.

Initially the answer of proposers agreed with me but later we had a slight different conclusions. As the per the proposers they arrived to $\frac{1}{\sqrt[\pi]{e^4}}$.

If the answer of proposers is correct then probably I think I had done mistakes while evaluating the limits. I even tried on wolfram alpha it agrees with me ( doesn't provide the closed form but it approximates to my closed form) . Can anybody help me to correct my mistakes ? Thank you ! :)

Solution Here i wish to share my solution.

Firstly we recall Euler's formula$e^{imx} = (\cos mx + i\sin mx)= (\cos x+ i\sin x)^m$ and further solving gives $\frac{\cos mx +i\sin mx}{\sin^m x}=$$\displaystyle \sum_{r=0}^m { m\choose r} \cot^{m-r}x i^{r}= \displaystyle \sum_{r=0}^{\lceil \frac{m}{2}\rceil+ l }(-1)^{r}{ m\choose 2r} \cot^{m-2r}x+i\sum_{r=1}^{\lceil \frac{m}{2}\rceil}{ m\choose 2r-1}(-1)^{r} \cot^{m-2r+1}x$ where $l=1,0$ if $m$ is even and odd respectively. We note that latter sum is the imaginary part and hence equating we get $\sum_{r=1}^{\lceil \frac{m}{2}\rceil}{ m\choose 2r-1} \cot^{m-2r+1}x=\frac{\sin mx}{\sin^{m}x}$ on setting $m= 2n+1, \; x= \frac{k\pi}{m}$ and further expansion we get ${ 2n+1 \choose 1} \cot^{2n} \frac{k\pi}{2n+1} - { 2n+1 \choose 3} \cot^{2n-2}\frac{k\pi}{2n+1} +\cdots + { 2n+1 \choose 2n+1} =0$ as the polynomial equation is of even degree and hence by Vieta's formula we have $\prod_{k=1}^{n} \cot^2\left(\frac{k\pi}{2n+1}\right)=(-1)^{2n}\frac{{ 2n+1\choose 2n+1}}{{ 2n+1\choose 1}}=\frac{1}{2n+1}$ secondly we use the identity $\displaystyle\prod_{k=0}^{n-1} \sin\left(x+\frac{k\pi}{n}\right) =\frac{\sin nx}{2^{n-1} }$, taking $\log$ on both side and on differentiating with respect to $x$, we get that, ie $\frac{d}{dx}\log\left(\displaystyle\prod_{k=0}^{n-1} \sin\left(x+\frac{k\pi}{n}\right)\right) =\frac{d}{dx}\log\left(\frac{\sin nx}{2^{n-1}}\right)\\ \Rightarrow \displaystyle \sum_{k=0}^{n-1} \cot\left(x+\frac{k\pi}{n}\right) =n\cot(nx)$. Replacing $n$ by $2n+1$ we yield $\displaystyle \sum_{k=0}^{2n} \cot\left(x+\frac{k\pi}{2n+1}\right) =(2n+1)\cot((2n+1)x)$and we deduce that $\omega_n = \left(\frac{(2n+1)\cot ((2n+1)x)}{2n+1}\right)^{-1} = \tan ((2n+1)x)$.

Call $L(n)= \displaystyle\lim_{x\to\frac{\pi}{4n+2}}\left(\frac{\tan x}{\tan \beta} \right)^{\omega_n}$. As we can observe that we have $1^{\infty}$ limit form. Since as soon as $x\to\frac{1}{4n+2}$,function $w_n\to\infty$ and $\frac{\tan x}{\tan\beta}\to 1$ with $\beta =\frac{\pi}{4n+2}$. So we can either make the direct use the formula for $1^{\infty}$ or without of it too. ie

$\displaystyle\lim_{x\to\beta}\left(\frac{\tan x}{\tan \beta}\right)^{\omega_n}=\displaystyle\lim_{x\to\beta}\left(1+\frac{\tan x}{\tan \beta}-1\right)^{\omega_n}= \displaystyle\lim_{x\to\beta}\left(1+\frac{1}{\frac{\tan x}{\tan \beta}-1}\right)^{\omega_n} \\= \exp\left(\displaystyle\lim_{x\to\beta }\left(\frac{\tan x}{\tan \beta }-1\right)\tan((2n+1)x)\right)=\exp\left(\displaystyle\lim_{x\to\beta }\left(\frac{\tan x-\tan\beta}{\tan\beta \cot ((2n+1)x)}\right)\right)$. Here we have the limit of the form $\frac{0}{0}$ so we use L-hopital's rule to get $\exp\left(\displaystyle \lim_{x\to\beta}\frac{\sec^2 x}{-(2n+1)\csc^2((2n+1)x)\tan\beta}\right)=\exp\left(\frac{\sec \beta}{-\sin\beta (2n+1)}\right)$. And therefore, $\lim_{n\to\infty} L(n)=\exp\left(-\displaystyle \lim_{n\to\infty} \frac{\frac{2\pi}{\pi(4n+2)}}{\tan \frac{\pi}{4n+2}}\right)=\exp\left(-\frac{2}{\pi}\displaystyle \lim_{n\to\infty} \frac{\frac{\pi}{(4n+2)}}{\tan \frac{\pi}{4n+2}}\right) =e^{-\frac{2}{\pi}}=\frac{1}{\sqrt[\pi]{e^{2}}}$