November 6th
Today I learned a proof of the Basel problem from Fourier analysis. As an example, we compute the Fourier series of the step function $s(x),$ which is $1$ if $\{x\} \lt \frac12$ and $-1$ otherwise. Our period is $[0,1),$ so our Fourier transform is\[\int_0^1s(x)e^{-2n\pi ix}\,dx=\int_0^{1/2}e^{-2n\pi ix}\,dx-\int_{1/2}^1e^{-2n\pi ix}\,dx=2\int_0^{1/2}e^{-2n\pi ix}\,dx.\]This evaluates to\[2\cdot\frac{e^{-2n\pi ix}}{-2n\pi i}\bigg|_0^{1/2}=\frac{1-(-1)^n}{n\pi i}.\]So our Fourier series is\[s(x)=\sum_{n=-\infty}^\infty\frac{1-(-1)^n}{n\pi i}e^{2n\pi ix}.\]Pairing off $n$ with $-n,$ we get $\frac{1-(-1)^n}{n\pi i}\left(e^{2n\pi ix}-e^{-2n\pi ix}\right)=2\cdot\frac{1-(-1)^n}{n\pi}\sin(2n\pi x).$ This vanishes for $n$ even, so we get to shift this to\[s(x)=\sum_{n=0}^\infty\frac4{(2n+1)\pi}\sin(2(2n+1)\pi x)=\frac4\pi\sum_{n=0}^\infty\frac{\sin(2(2n+1)\pi x)}{2n+1}.\]
We are actually almost done because I don't care about rigor. Integrating from $0$ to $\frac12,$ the step function accumulates a full $\frac12.$ But on the other side,\[\int_0^{1/2}\sin(2n\pi x)\,dx=\frac{-\cos(2n\pi x)}{2n\pi}\bigg|_0^{1/2}=\frac1{n\pi}.\]So taking the integral on both sides of our Fourier series, we see\[\frac12=\frac4\pi\sum_{n=0}^\infty\frac1{(2n+1)^2\pi}.\]This rearranges to\[\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8,\]which solves the Basel problem. Explicitly, we can write\[\frac34\sum_{n=0}^\infty\frac1{n^2}=\sum_{n=0}^\infty\frac1{n^2}-\frac1{2^2}\sum_{n=0}^\infty\frac1{n^2}=\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8.\]This rearranges into $\sum_n\frac1{n^2}=\frac{\pi^2}6,$ as desired.