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?

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
${2n \choose 2n-2} = 2n^2-n.$
