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