Another simple solution to the Basel problem

Recall that $\zeta(2)=\sum_{k=1}^\infty 1/k^2$ . Perhaps the shortest proof of $\zeta(2)=\pi^2/6$ applies (the sophisticated) Parseval’s identity of the Fourier series to $f(x)=x$ . By contrast, the simple (but long) proof in this paper, which was recently popularized by the video below, uses basic ideas from Euclidean geometry.

The following argument interpolates between these two, resulting in a proof that is both simple and short.

Take $n$ even, put $[n]:=\{0,\ldots,n-1\}$ , and define $h\colon[n]\rightarrow\mathbb{C}$ by $h(j):=1$ for $j<n/2$ and $h(j):=0$ otherwise. Then the geometric sum formula gives

$\displaystyle \sum_{j\in[n]}h(j)e^{2\pi ijk/n}=\left\{\begin{array}{ll}n/2&\text{if }k\equiv0\bmod n\\2/(1-e^{2\pi ik/n})&\text{if }k\text{ odd}\\0&\text{otherwise.}\end{array}\right.$

Next, we apply Parseval’s identity of the discrete Fourier transform, whose proof also follows from the geometric sum formula:

$\displaystyle 4\sum_{\substack{k\in[n]\\k\text{ odd}}}\frac{1}{|1-e^{2\pi ik/n}|^2}=\sum_{k\in[n]}\bigg|\sum_{j\in[n]}h(j)e^{2\pi ijk/n}\bigg|^2-\Big(\frac{n}{2}\Big)^2$

$\displaystyle =n\sum_{j\in[n]}|h(j)|^2-\Big(\frac{n}{2}\Big)^2=\frac{n^2}{4}.$

(An alternative geometric proof of this identity appears in Wastlund’s paper for the case where $n$ is a power of 2.) For $n\gg k$ , we have $e^{2\pi ik/n}-1\approx 2\pi ik/n$ , suggesting we select an increasing function $K(n)$ so that

$\displaystyle S:=\sum_{\substack{k=-\infty\\k\text{ odd}}}^\infty\frac{1}{k^2}=\lim_{n\rightarrow\infty}\sum_{\substack{|k|\leq K(n)\\k\text{ odd}}}\frac{1}{k^2}=\lim_{n\rightarrow\infty}\sum_{\substack{|k|\leq K(n)\\k\text{ odd}}}\underbrace{\frac{1}{|1-e^{2\pi ik/n}|^2}\cdot\Big(\frac{2\pi}{n}\Big)^2}_{a_{k,n}}=\frac{\pi^2}{4}.$

(In the last sum, we take odd $k\in\{-\lfloor K(n)\rfloor,\ldots,\lfloor K(n)\rfloor\}$ modulo $n$ .) Provided this holds, then we see from the terms of $S$ that $\zeta(2)=S/2+4\zeta(2)$ , from which it follows that $\zeta(2)=\pi^2/6$ .

To prove $S=\pi^2/4$ , we bound the truncation error for any $K\ll n$ :

$\displaystyle \bigg|\sum_{\substack{|k|\leq K\\k\text{ odd}}}\frac{1}{k^2}-\frac{\pi^2}{4}\bigg|=\sum_{\substack{|k|\leq K\\k\text{ odd}}}\bigg(a_{k,n}-\frac{1}{k^2}\bigg)+\sum_{\substack{|k|>K\\k\text{ odd}}}a_{k,n}$

$\displaystyle \leq n\bigg(a_{K,n}-\frac{1}{K^2}\bigg)+na_{K,n}\leq 2na_{K,n}\leq \frac{4n}{K^2},$

where the last step applies the fact that $\cos x\leq 1-x^2/2+x^4/24$ and assumes $K\leq n/3$ . As such, we may take $K=K(n)=n^{2/3}$ (say) to obtain the desired convergence.

	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…
	Jaan Parts on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…
	Tom Sirgedas on Polymath16, seventeenth thread…

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