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? $$\sum_{k=1}^{n}\frac{1}{2-\cos \frac{2\pi k}{2n+1}}=?$$

Solution 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).