Proof verification: $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$
Solution 1:
To make my comments' strategy explicit, $\arcsin t=\sum_{n\ge0}\frac{(2n-1)!!}{(2n)!!(2n+1)}t^{2n+1}$ for $|t|\le1$, and$$\int_0^{\pi/2}\sin^{2n+1}xdx=\tfrac12\operatorname{B}(n+1,\,\tfrac12)=\frac{n!\sqrt{\pi}}{2\Gamma(n+\tfrac32)}=\frac{n!2^n}{(2n+1)!!}=\frac{(2n)!!}{(2n+1)!!},$$so$$\begin{align}\frac{\pi^2}{8}&=\int_0^{\pi/2}xdx\\&=\int_0^{\pi/2}\arcsin\sin xdx\\&=\sum_{n\ge0}\frac{(2n-1)!!}{(2n)!!(2n+1)}\int_0^{\pi/2}\sin^{2n+1}xdx\\&=\sum_{n\ge0}\frac{1}{(2n+1)^2}.\end{align}$$