Proving Summation: $\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}

[Mentallic]

$$\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$$

I'd like to know how to prove this summation. And if possible, what is the significance of having $\pi$ in the answer?

 

[uart]

If you're familar with Fourier series then one nice method is to consider the expansion of a "saw-tooth" wave as follows.

Let $y(x) = \pi x$ : $-0.5 \leq x \leq 0.5$

Now make it periodic as per $y(x) = y(x-k)$ : $-0.5+k \leq x \leq 0.5+k$, for all integer k.

It's fairly easy to show that the Fourier series expansion is,

$$y = \sin(2 \pi x) - \frac{1}{2} \sin(4 \pi x) \, ... \, + \frac{(-1)^{k+1}}{k} \sin(2k \pi x) + \, ...$$

Consider the mean squared value of y, calculated two different ways. Firstly calculate directly from y(x),

$$MS(y) = \int_{x=-0.5}^{+0.5} (\pi x)^2 dx = \frac {\pi^2}{12}$$

Now repeating the calculation but this time using the Fourier series (and making use of the fact that the terms are orthagonal) we get,

$$MS(y) =0.5 ( 1 + 1/4 + 1/9 + ... 1/k^2 + ... )$$

Equating these two expressions for the mean squared value gives the required sum.

 

[g_edgar]

Here is Robin Chapman's collection of 14 different proofs that zeta(2) = pi^2/6 ...

http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf