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Please help me to prove that

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

Thanks.

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    $\begingroup$ Duplicate: math.stackexchange.com/questions/562360/… $\endgroup$ – Winther Sep 21 '16 at 9:22
  • $\begingroup$ There is a particular polynomial $p_N(x)$ having $\cos\frac{\pi k}{N}$ for $k=1,2,\ldots,N$ as roots. Apply Vieta's formulas to $q_N(x)=p_N(1-x)$ in order to find the sum of the reciprocal of the roots of $q_N(x)$. $\endgroup$ – Jack D'Aurizio Sep 21 '16 at 22:56
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    $\begingroup$ Ilham: DO. NOT. Reask the same question. This is a rule that is enforced strictly. Edit this first version to shape, if you are not happy with the way you wrote it. If you accidentally created two accounts, then read this. $\endgroup$ – Jyrki Lahtonen Sep 22 '16 at 6:55
  • $\begingroup$ @JackD'Aurizio This isn't your first rodeo. You can check the links Winther gave, and see that this was a repost by the same asker (in addition to being essentially a duplicate of an earlier one). Anyway, I'm merging and deleting the reposted thread. $\endgroup$ – Jyrki Lahtonen Sep 22 '16 at 6:56
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    $\begingroup$ We may notice that $\cos\frac{\pi k}{N}$ for $k=1,2,\ldots,N-1$ are the roots of the [Chebyshev polynomial][1] $U_{N-1}(x)$. 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 $q_N(x)$.<br> Since the Chebyshev polynomials fulfill the recurrence relation $$ U_{N}(x) = 2x\cdot U_{N-1}(x)-U_{N-2}(x) $$ $\endgroup$ – Jack D'Aurizio Sep 22 '16 at 16:04
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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.


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.


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 $1-\cos\theta = 2\cos^2\frac{\theta}{2}$.

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