<s>[INST] Roots of a polynomial whose coefficients are ratios of binomial coefficients 
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots +  \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$
Hence prove that
$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$

I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.
Any help will be appreciated.
Thanks.
 [/INST] We have
$$\frac{e^{m i \theta}}{\sin^m \theta} = \frac{(\cos \theta + i \sin \theta)^m}{\sin^m \theta} = (\cot \theta + i)^m $$
for $m \in \mathbb{N}$ and $\theta \in \mathbb{R}$. Take $m=2n+1$ and $\theta = k\pi/(2n+1)$ to get
$$\frac{(-1)^k}{\sin^{2n+1}\theta} = \sum_{r=0}^{2n+1} \binom{2n+1}{r}i^r \cot^{2n+1-r} \theta . $$
The imaginary part comes from the summands for which $r$ is odd, hence
$$0 = \sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s (\cot^2\theta)^{n-s}. $$
Equivalently, the polynomial
$$\sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s x^{n-s} $$
has $\cot^2 \theta$ as a root. Divide by $2n+1$ to get the relation in the question. For the second part, take the coefficient of $x^{n-1}$ to get
$$\sum_{k=0}^n (\cot \frac{k\pi}{2n+1})^2 = \frac{1}{2n+1}\binom{2n+1}{3}. $$
Now divide by $2n(2n-1)$ and take a limit using $\cot \theta = 1/\theta + O(\theta)$.
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