<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] I'll post the one I know since it is Euler's, and is quite easy and stays in $\mathbb{R}$. (I'm guessing Euler didn't have tools like residues back then).
Let
$$s = {\sin ^{ - 1}}x$$
Then
$$\int\limits_0^{\frac{\pi }{2}} {sds}  = \frac{{{\pi ^2}}}{8}$$
But then 
$$\int\limits_0^1 {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}  = \frac{{{\pi ^2}}}{8}$$
Since
$${\sin ^{ - 1}}x = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}  = x + \frac{1}{2}\frac{{{x^3}}}{3} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5} + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{{{x^7}}}{7} +  \cdots $$
We have
$$\int\limits_0^1 {\left\{ {\frac{{dx}}{{\sqrt {1 - {x^2}} }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} } \right\}}  = \int\limits_0^1 {\left\{ {x + \frac{1}{2}\frac{{{x^3}}}{3}\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5}\frac{{dx}}{{\sqrt {1 - {x^2}} }} +  \cdots } \right\}} $$
But
$$\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 - {x^2}} }}dx}  = \frac{{2n}}{{2n + 1}}\int\limits_0^1 {\frac{{{x^{2n - 1}}}}{{\sqrt {1 - {x^2}} }}dx} $$
which yields
$$\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 - {x^2}} }}dx}  = \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}$$
since all powers are odd.
This ultimately produces:
$$\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{2}\frac{1}{3}\left( {\frac{2}{3}} \right) + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{1}{5}\left( {\frac{{2 \cdot 4}}{{3 \cdot 5}}} \right) + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{1}{7}\left( {\frac{{2 \cdot 4 \cdot 6}}{{3 \cdot 5 \cdot 7}}} \right) \cdots $$
$$\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} +  \cdots $$
Let
$$1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} +  \cdots  = \omega $$
Then
$$\frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} +  \cdots  = \frac{\omega }{4}$$
Which means
$$\frac{\omega }{4} + \frac{{{\pi ^2}}}{8} = \omega $$
or
$$\omega  = \frac{{{\pi ^2}}}{6}$$
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