<s>[INST] The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
 [/INST] Another variation. We make use of the following identity (proved at the bottom of this note):
$$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. \quad (1)$$
Now $1/\theta > \cot \theta > 1/\theta - \theta/3 > 0$ for $0< \theta< \pi/2 < \sqrt{3}$ and so
$$ 1/\theta^2 – 2/3 < \cot^2 \theta < 1/\theta^2. \quad (2)$$
With $\theta_k = (2k-1)\pi/4n,$ summing the inequalities $(2)$ from $k=1$ to $n$ we obtain   
$$2n^2 – n < \sum_{k=1}^n \left( \frac{2n}{2k-1}\frac{2}{\pi} \right)^2 < 2n^2 – n + 2n/3.$$
Hence
$$\frac{\pi^2}{16}\frac{2n^2-n}{n^2} < \sum_{k=1}^n \frac{1}{(2k-1)^2} <
\frac{\pi^2}{16}\frac{2n^2-n/3}{n^2}.$$
Taking the limit as $n \rightarrow \infty$ we obtain
$$  \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{\pi^2}{8},$$
from which the result for $\sum_{k=1}^\infty 1/k^2$ follows easily.
To prove $(1)$ we note that
$$ \cos 2n\theta = \text{Re}(\cos\theta + i \sin\theta)^{2n} = 
\sum_{k=0}^n (-1)^k {2n \choose 2k}\cos^{2n-2k}\theta\sin^{2k}\theta.$$
Therefore
$$\frac{\cos 2n\theta}{\sin^{2n}\theta} =  \sum_{k=0}^n (-1)^k {2n \choose 2k}\cot^{2n-2k}\theta.$$
And so setting $x = \cot^2\theta$ we note that
$$f(x) = \sum_{k=0}^n (-1)^k {2n \choose 2k}x^{n-k}$$
has roots $x_j = \cot^2 (2j-1)\pi/4n,$ for $j=1,2,\ldots,n,$ from which $(1)$ follows since
${2n \choose 2n-2} = 2n^2-n.$
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