The Basel problem regards one of my favorite infinite series.

It asks what is the value of $$ \sum_{k=1}^{\infty} \frac{1}{k^2} \:. $$ Of course we now know the answer to be $\frac{\pi^2}{6}$, but this problem stumped many mathematicians back in the day. Euler was the first to propose the correct solution, but it took many years later until his argument was made rigourous. $ \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\abs}[1]{| #1 |} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ind}{\mathbf{1}} \newcommand{\ip}[2]{\langle #1, #2 \rangle} $

I think one of the cleanest proofs (in the sense of the least amount of calculations needed) is based on a very simple Fourier series argument, which I will describe in this post. I like this proof because it demonstrates the power of Fourier analysis on a very concrete problem.

We will work with the function space $L^2([0, 1])$. Consider the function $L^2([0, 1]) \ni f : [0, 1] \longrightarrow \R$ defined as $f(x) = 1/2 - x$. The Fourier coefficients of $f$ are given by $$ \widehat{f}(k) = \int_0^1 f(x) e^{-2\pi j k x} \; dx \:,\: k \in \Z \:. $$ Immediate we can see that $\widehat{f}(0) = 0$, and a simple integration by parts argument yields that $$ \widehat{f}(k) = \frac{1}{2\pi j k} \:, \: k \neq 0, k \in \Z \:. $$

Now comes the fun part. The essential fact about Fourier series on $L^2([0, 1])$ is that the characters $$ \{ \phi_k(x) = e^{-2\pi j k x} : k \in \Z \} $$ form an orthonormal basis for $L^2([0, 1])$. Hence, Parseval's identity gives us that $$ \norm{ f }_{L^2([0,1])}^2 = \sum_{k \in \Z} \abs{\ip{f}{\phi_k}_{L^2([0,1])}}^2 = \sum_{k\in Z} \abs{\widehat{f}(k)}^2 = \frac{2}{4\pi^2} \sum_{k=1}^{\infty} \frac{1}{k^2} \:. $$ This is the essential step: the term we want appears on the right hand side, and to finish the argument we just need to compute the norm of $f$. But this is another straightforward integral $$ \norm{f}^2_{L^2([0,1])} = \int_0^1 (1/2-x)^2 \; dx = \frac{1}{12} \:. $$ Now it is just a matter of algebra. $$ \frac{1}{12} = \frac{2}{4\pi^2} \sum_{k=1}^{\infty} \frac{1}{k^2} \Longrightarrow \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \:. $$ And there we have it.

P.S. The proof presented here is well known, but the particular instance of $f$ and orthonormal basis chosen is from Folland's Real Analysis, Chapter 8. Actually, Folland regards $f$ as a periodic function mapping the torus $\mathbb{T} = \R / \Z$ into $\R$, but the approach above is identical.

P.P.S. Thanks to Mark Dominus for pointing out to me that Basel is a name of a city in Switzerland, and not the name of a person as I had originally thought.