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

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

Thanks.

• Sep 21, 2016 at 9:22
• 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 $q_N(x)$. Sep 21, 2016 at 22:56
• 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. Sep 22, 2016 at 6:55
• @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. Sep 22, 2016 at 6:56
• 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 $q_N(x)$.<br> Since the Chebyshev polynomials fulfill the recurrence relation $$U_{N}(x) = 2x\cdot U_{N-1}(x)-U_{N-2}(x)$$ Sep 22, 2016 at 16:04

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