While milking the integral $\int_0^\pi\sin^{-1}\left(\sin x\right)dx$, I believe I may have conceived of a nice proof of $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ It makes use of the established fact that an equivalent problem is to prove that $$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
Proof: since $\sum 1/n^2$ and $\sum 1/(2n+1)^2$ both converge, the chain of equalities \begin{align} \sum_{n=1}^{2k}\frac{1}{n^2} &= 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{(2k-1)^2}+\frac{1}{(2k)^2}\\ &= 1+\frac{1}{3^2}+\cdots+\frac{1}{(2(k-1)+1)^2}+\frac{1}{2^2}+\frac{1}{4^2}+\cdots+\frac{1}{(2k)^2}\\ &= \sum_{n=0}^{k-1}\frac{1}{(2n+1)^2}+\sum_{n=1}^k\frac{1}{(2n)^2}\\ &= \sum_{n=0}^{k-1}\frac{1}{(2n+1)^2}+\frac{1}{4}\sum_{n=1}^k\frac{1}{n^2} \end{align} implies that $\sum_{n=1}^\infty 1/n^2=\frac{4}{3}\sum_{n=0}^\infty 1/(2n+1)^2$, so $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}\iff\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
I'd greatly appreciate it if someone took the time to review it!
Proof: consider the integral $$\int_0^\pi\sin^{-1}\left(\sin x\right)dx$$ On one hand, \begin{align} \int_0^\pi\sin^{-1}\left(\sin x\right)dx &= \int_0^{\frac{\pi}{2}}\sin^{-1}\left(\sin x\right)dx+\int_{\frac{\pi}{2}}^\pi\sin^{-1}\left(\sin x\right)dx\\ &= \int_0^{\frac{\pi}{2}}xdx+\int_{\frac{\pi}{2}}^\pi(\pi-x)dx\\ &= \frac{1}{2}\left(\frac{\pi}{2}\right)^2+\pi\left(\pi-\frac{\pi}{2}\right)-\left[\frac{1}{2}\pi^2-\frac{1}{2}\left(\frac{\pi}{2}\right)^2\right]\\ &= \frac{\pi^2}{8}+\pi^2-\frac{\pi^2}{2}-\frac{\pi^2}{2}+\frac{\pi^2}{8}\\ &= \frac{\pi^2}{4} \end{align} On the other, the fact that $-1\leq\sin(x)\leq 1$ and $\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}t^{2n+1}$ converges uniformly to $\sin^{-1}(t)$ over $[-1,1]$ as $k\to\infty$ justifies us writing \begin{align} \int_0^\pi\sin^{-1}\left(\sin x\right)dx &= \lim_{k\to\infty}\int_0^\pi\left(\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\sin^{2n+1}(x)\right)dx\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\int_0^\pi\sin^{2n+1}(x)dx\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\left(\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx+\int_{\frac{\pi}{2}}^\pi\sin^{2n+1}(x)dx\right)\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\left(\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx+\int_{\frac{\pi}{2}-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}}\sin^{2n+1}\left(x+\frac{\pi}{2}\right)dx\right)\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\left(\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx+\int_0^{\frac{\pi}{2}}\cos^{2n+1}(x)dx\right)\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\left(\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx+\int_{\frac{\pi}{2}-0}^{\frac{\pi}{2}-\frac{\pi}{2}}\cos^{2n+1}\left(\frac{\pi}{2}-x\right)dx\right)\\ &= \lim_{k\to\infty}\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\left(\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx+\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx\right)\\ &= \lim_{k\to\infty}2\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx\\ \end{align} All that remains is to evaluate $\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx$. Using the reduction formula for $\int\sin^n (x)dx$, it can be shown that for every integer $n\geq 1$, $$\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx=\frac{2n}{2n+1}\int_0^{\frac{\pi}{2}}\sin^{2n-1}(x)dx$$ and, after inducting on $n$, \begin{align} \int_0^{\frac{\pi}{2}}\sin^{2n+1}(x)dx &= \int_0^{\frac{\pi}{2}}\sin(x)dx\cdot\frac{2(1)}{2(1)+1}\cdot\frac{2(2)}{2(2)+1}\cdots\frac{2n}{2n+1}\\ &= \frac{2\cdot 4\cdots (2n)}{1\cdot 3\cdots (2n+1)}\\ &= \frac{2^n(1\cdot 2\cdots n)}{1\cdot 3\cdots(2n+1)}\\ &= \frac{2^n(1\cdot 2\cdots n)\cdot(2\cdot 4\cdots(2n))}{1\cdot 2\cdot 3\cdot 4\cdots(2n)\cdot(2n+1)}\\ &= \frac{2^n(1\cdot 2\cdots n)\cdot 2^n(1\cdot 2\cdots n)}{(2n+1)!}\\ &= \frac{\left(2^n n!\right)^2}{(2n+1)!}\\ \end{align} Substituting this expression into our limit, we get \begin{align} \int_0^\pi\sin^{-1}\left(\sin x\right)dx &= \lim_{k\to\infty}2\sum_{n=0}^k\frac{(2n)!}{(2^n n!)^2(2n+1)}\cdot\frac{\left(2^n n!\right)^2}{(2n+1)!}\\ &= \lim_{k\to\infty}2\sum_{n=0}^k\frac{(2n)!}{(2n+1)}\cdot\frac{1}{(2n)!(2n+1)}\\ &= \lim_{k\to\infty}2\sum_{n=0}^k\frac{1}{(2n+1)^2}\\ &= 2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\\ \end{align} which, together with the original result $\int_0^\pi\sin^{-1}\left(\sin x\right)dx=\frac{\pi^2}{4}$, immediately yields $$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{1}{2}\cdot\frac{\pi^2}{4}=\frac{\pi^2}{8}$$ Q.E.D.
Let me know what you think! If you identify any errors or optimizations, please don't hesitate to share them with me.
