Finding $\sum_{k=1}^{n}\frac{1}{2-\cos \frac{2\pi k}{2n+1}}$ using the cotangent function I found the sum of this series using the $ \cot x $ function:
$$\sum_{k=1}^{n}\frac{1}{1-\cos \frac{2\pi k}{2n+1}}=\frac{1}{2}\sum_{k=1}^n\frac{1}{\sin^2(\frac{k\pi}{2n+1})}=\frac{1}{2}\sum_{k=1}^n\left(1+\cot^2\left(\frac{k\pi}{2n+1}\right)\right)$$
$$=\frac{n}{2}+\frac{1}{2}\sum_{k=1}^n\cot^2\left(\frac{k\pi}{2n+1}\right)=\frac{n(n+1)}{3}$$

Question: Is it possible to find the sum of a new series using the $\cot x$ function?
$$\sum_{k=1}^{n}\frac{1}{2-\cos \frac{2\pi k}{2n+1}}=?$$


You might as well solve it for a free variable, which is performed now.
From Hansen, Table of Series and Products, entry 91.2.10 specialized to odd integers,
$$\prod_{k=1}^n \cosh{y} - \cos{ \big(\frac{2 \pi k}{2n+1} \big)} = 2^{-n} 
\frac{\sinh{\big((n+1/2)y\big)}}{\sinh{\big(y/2\big)}} $$
Do a logarithmic derivative on both sides and simplify:
$$ \sum_{k=1}^n \frac{1}{ \cosh{y} - \cos{ \big(\frac{2 \pi k}{2n+1} }\big) } =
\frac{1}{\sinh{y}}\Big( (n+1/2)\coth{\big((n+1/2)y\big)}  -  1/2\coth{\big(y/2\big)} \Big)$$
The OP's solution is for $y=$arccosh(2).
