How to prove that $\sum _{k=1} ^N \frac 1 {1-\cos \frac {k\pi} N} = \frac {2N^2 + 1} 6$? [closed]

Question

Please help me to prove that

$$\sum _{k=1} ^N \frac 1 {1-\cos \frac {k\pi} N} = \frac {2N^2 + 1} 6 .$$

Thanks.

Answer 1

We may notice that $\cos\frac{\pi k}{N}$ for $k=1,2,\ldots,N-1$ are the roots of the Chebyshev polynomial $U_{N-1}(x)$. If we define $$ q_N(x) = U_{N-1}(1-x) $$ then the given sum is $\frac{1}{2}$ plus the sum of the reciprocal of the roots of $q_N(x)$.
Since the Chebyshev polynomials fulfill the recurrence relation $$ U_{N}(x) = 2x\cdot U_{N-1}(x)-U_{N-2}(x) $$ it is not difficult to prove by induction on $N$ that $$ q_N(x)=U_{N-1}(1-x) = (-1)^{N-1}2^{N-1}x^{N-1}+\ldots-\color{green}{\frac{N(N^2-1)}{3}}x+\color{blue}{N} $$ hence by Vieta's formulas $$ \sum_{k=1}^{N}\frac{1}{1-\cos\frac{\pi k}{N}}=\frac{1}{2}+\frac{\color{green}{N(N^2-1)}}{\color{green}{3}\color{blue}{N}} = \color{red}{\frac{2N^2+1}{6}}$$ a wanted.

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You may also avoid induction by noticing that the critical ratio between the opposite of the coefficient of $x$ and the coefficient of $1$ in $q_N(x)$ is $$ -\frac{q_N'(0)}{q_N(0)}=\lim_{x\to 0}\frac{(N-1) (1-x) U_{N-1}(1-x)-N U_{N-2}(1-x)}{\left(-1+(1-x)^2\right) U_{N-1}(1-x)}$$ or, by replacing $x$ with $1-\cos\theta$, $$ \lim_{\theta\to 0}\frac{(N-1) \cos(\theta) U_{N-1}(\cos\theta)-N U_{N-2}(\cos\theta)}{\left(-1+\cos^2\theta\right) U_{N-1}(\cos\theta)}$$ that equals $$ \lim_{\theta\to 0}\frac{(N-1) \cos(\theta)\sin(N\theta)-N\sin((N-1)\theta)}{-\sin^2(\theta)\sin(N\theta)}$$ and can be computed from the Taylor series of $\sin(\theta)$ and $\cos(\theta)$ at $\theta=0$, up to the $\theta^3$ term.

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Cauchy's proof of the identity $\zeta(2)=\frac{\pi^2}{6}$ also provides a technique for evaluating the given sum, once we realize that $1-\cos\theta = 2\cos^2\frac{\theta}{2}$.