Let gooby do teh hoemwerk: Riemann zeta at two

Based on what I read in a very old book, we show

\zeta(2) = \sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}

We can divide the sum into even and odd contributions

\sum\limits_{n=1}^\infty \frac{1}{n^2} = \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum\limits_{n=1}^\infty \frac{1}{(2n)^2}

One sees that

\zeta(2) =\frac{4}{3}\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}

In order to evaluate the right hand side of this equation we consider the function

C(x) = \sum\limits_{n=1}^\infty \frac{\cos (2n-1)x}{(2n-1)^2}, such that \zeta(2) =\frac{4}{3}C(0)

We now take a small detour and consider the finite sum C_N(x) and its derivatives

C_N(x) = \sum\limits_{n=1}^N \frac{\cos (2n-1)x}{(2n-1)^2}

C_N'(x) = -\sum\limits_{n=1}^N \frac{\sin (2n-1)x}{2n-1}

C_N''(x) = -\sum\limits_{n=1}^N \cos(2n-1)x

Multiplying the last equation by 2\sin x and using

2\sin x \cos k x = \sin(k+1)x-\sin(k-1)x

We get

2\sin x \,C_N''(x) = -\sin 2Nx

Now one can solve for C_N'(x)

C_N'(x) = C_N'(\pi/2) - \int\limits_{\pi/2}^{x}dy\,\frac{\sin 2Ny}{\sin y}

By writing

C_N'(x) = C_N'(\pi/2) + \frac{1}{2N} \int\limits_{\pi/2}^{x}dy\,\frac{\frac{d}{dy}\cos 2Ny}{\sin y}

One can think of integration by parts and notice that the second term vanishes when N\to\infty. Now

C'(\pi/2) = -\lim\limits_{N\to\infty}\sum\limits_{n=1}^N \frac{(-1)^n}{2n-1} =-\arctan(1) = -\frac{\pi}{4}

In other words,

C'(x) =-\frac{\pi}{4}

from which it is immediate

C(x) = C(0) - \frac{\pi}{4}x

Now note C(\pi/2) = 0 and therefore

C(0) = \frac{\pi}{4}\times\frac{\pi}{2} = \frac{\pi^2}{8}

Finally, the identity \zeta(2) =\frac{4}{3}C(0) establishes the result \zeta(2) =\frac{\pi^2}{6}.

Happy new year to evri1!


About goobypl5

pizza baker, autodidact, particle physicist
This entry was posted in Math and tagged , , , . Bookmark the permalink.

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