Boming Jia

The University of Chicago
Department of Mathematics
Eckhart Hall

jiab@uchicago.edu

Zeta(2)

This article is an exposition about $$\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{\pi ^2}{6}$$ It is not easy to find an elementary and easy-to-remember-yet-still-rigorous proof of this theorem. However, recently I have encountered such a proof in a Monthly paper by Josef Hofbauer.

Here is a link to the paper:

http://homepage.univie.ac.at/Josef.Hofbauer/02amm.pdf

First let's review some basic trigonometry.

Given any angle $x$, we have ${\sin^2 x}+\cos^2x=1$. Then we have the following identity, which could be checked easily by drawing a picture. $$\sin(2x) = 2 \sin x \cos x$$

Also by looking at the graph of $\sin$ and $\cos$, we can easily check the following: \begin{align*} \cos x & = \sin (x+\frac{\pi }{2}) \\ \sin x & =\sin(\pi -x) \end{align*}

Now we start the proof.

Since $\sin(2x) = 2 \sin x \cos x$, we have $$\sin x= 2 \sin(\frac{x}{2})\cos(\frac{x}{2}).$$

Taking the reciprocals and squaring both sides, we get $$\frac{1}{\sin^2 x} = \frac{1}{4}\frac{1}{\sin^2(x/2)\cos^2(x/2)}.$$

Then by ${\sin^2 x}+\cos^2x=1$, we have $$\frac{1}{\sin^2 x} = \frac{1}{4}\frac{\sin^2 (x/2)+\cos^2 (x/2)}{\sin^2(x/2)\cos^2(x/2)}=\frac{1}{4}( \frac{1}{\cos^2(x/2)} + \frac{1}{\sin^2(x/2)}).$$ Now applying $\cos(x/2) = \sin((x+\pi )/2)$, we get $$\ \ \ \ \ \ \frac{1}{\sin^2 x} = \frac{1}{4} (\frac{1}{\sin^2\frac{x}{2}} + \frac{1}{\sin^2\frac{x+\pi }{2}}). \ \ \ \ \ \ \ \ \ \ (*) $$ This is the crucial step in the entire proof, let call this indentity (*).

According to the definition of $\sin$,$$\sin(\pi /2)=\,\sin90^\circ =1.$$ Then applying (*) multiple times, we have \begin{align*} 1 & = \frac{1}{\sin^2(\pi /2)} \\ & =\frac{1}{4} (\frac{1}{\sin^2(\pi /4)} + \frac{1}{\sin^2(3\pi /4)}) \\ & =\frac{1}{4^2} (\frac{1}{\sin^2(\pi /8)} + \frac{1}{\sin^2(3\pi /8)}+\frac{1}{\sin^2(5\pi /8)} + \frac{1}{\sin^2(7\pi /8)})\\ & = \dots \end{align*} And we can keep doing this.

Now apply the identity $\sin(\pi -x)=\sin x$, we get \begin{align*} 1 & =\frac{2}{4^2} (\frac{1}{\sin^2(\pi /8)} + \frac{1}{\sin^2(3\pi /8)}) \\ & =\frac{2}{4^3} (\frac{1}{\sin^2(\pi /16)} + \frac{1}{\sin^2(3\pi /16)}+\frac{1}{\sin^2(5\pi /16)} + \frac{1}{\sin^2(7\pi /16)}) \\ & ={\frac{2}{4^4} (\frac{1}{\sin^2(\pi /32)} + \frac{1}{\sin^2(3\pi /32)}+\frac{1}{\sin^2(5\pi /32)} +\dots + \frac{1}{\sin^2(15\pi /32)})}\\ & = \dots \end{align*} Let's denote this equation by (**)

Some readers might want to ask, why we need to do the last step and get (**)?

The answer is: in (**) all the angles in $\sin()$ are acute, and for an acute $x$,we have $$\sin x < x < \tan x$$

Taking the reciprocals and squaring each terms, we get $$\frac{1}{\sin^2 x} > \frac{1}{x^2} > \frac{1}{\tan^2 x}.$$

And we also know that $$\frac{1}{\tan^2 x}=\frac{\cos^2 x}{\sin^2 x}=\frac{1-\sin^2 x}{\sin^2 x}=\frac{1}{\sin^2 x}-1.$$

So \begin{align}\frac{1}{\sin^2 x}-1<\frac{1}{x^2}<\frac{1}{\sin^2 x}.\end{align}

Now together with equation (**): \begin{align}1= \frac{2}{4^4} (\frac{1}{\sin^2(\pi /32)} + \frac{1}{\sin^2(3\pi /32)}+\frac{1}{\sin^2(5\pi /32)} +\dots + \frac{1}{\sin^2(15\pi /32)}).\end{align}

We could get the following inequalities: \begin{align} 1-\frac{2}{4^4}*2^3 & <{\frac{2}{4^4} (\frac{1}{(\pi /32)^2} + \frac{1}{(3\pi /32)^2}+\frac{1}{(5\pi /32)^2} +\dots + \frac{1}{(15\pi /32)^2})} < 1,\\ 1-\frac{2}{4^4}*2^3 & < {\frac{2}{4^4}* 4^5 (\frac{1}{\pi ^2} + \frac{1}{(3\pi )^2}+\frac{1}{(5\pi )^2} +\dots + \frac{1}{(15\pi )^2})} <1,\\ 1-\frac{1}{2^4}& < {\ 8\,(\frac{1}{\pi ^2} + \frac{1}{(3\pi )^2}+\frac{1}{(5\pi )^2} +\dots + \frac{1}{(15\pi )^2})} <1. \end{align}

By examine what we have done carefully, we notice that in the sum of (**) we could get more terms than $$1 = {\frac{2}{4^4} (\frac{1}{\sin^2(\pi /32)} + \frac{1}{\sin^2(3\pi /32)}+\frac{1}{\sin^2(5\pi /32)} +\dots + \frac{1}{\sin^2(15\pi /32)})}.$$ In general if we use apply (*) multiple times, we can get $2^(n-1)$ terms of $\sin$ in (**).

Then we have $$1-\frac{1}{2^n} < {8(\frac{1}{\pi ^2} + \frac{1}{(3\pi )^2}+\frac{1}{(5\pi )^2} +\cdots + \frac{1}{((2^{n}-1)\pi )^2})}<1.$$

As $n$ becomes large enough (goes to infinity), $1/2^{n}$ can be ignored (converges to zero), so we have $${\,8\,(\frac{1}{\pi ^2} + \frac{1}{(3\pi )^2}+\frac{1}{(5\pi )^2}+\frac{1}{(7\pi )^2} +\cdots)}= 1,$$ i.e. $$\frac{1}{1^2}+\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +\cdots =\frac{\pi ^2}{8}. $$

Now we have almost finished proof.

Let $$\zeta(2) = \frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ ,$$

Then $$ \frac{\zeta(2)}{4} = \frac{1}{2^2}+\frac{1}{4^2} +\frac{1}{6^2} +\frac{1}{8^2} +\frac{1}{10^2} +\cdots\ .$$

Substract one from the other, we have $$\zeta(2)-\frac{\zeta(2)}{4} = \frac{1}{1^2}+\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +\cdots =\frac{\pi^2}{8}.\ \ \ \ \ \ \ \ \ \ \ \ $$

So $3\zeta(2) /4 = \pi ^2/8$, hence $\zeta(2) = \pi ^2/6$, i.e. $$\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{\pi ^2}{6}.$$

Q.E.D.

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