# Estimating the sum of reciprocal powers using a given fourier series

1. Oct 15, 2008

### buffordboy23

1. The problem statement, all variables and given/known data

Let f(x) be defined by the following Fourier series for $$\left|x\right|$$:

$$f(x) = \frac{\pi}{2} - \frac{4}{\pi}\sum_{1,3,...}\frac{cos\left(nx\right)}{n^{2}}$$

Show that

$$\sum_{1,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{8}$$

and

$$\sum_{1,2,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$$

3. The attempt at a solution

I was able to find the first sum by letting x = 0. I don't now how to approach the second part since the sum consists of the odd and even n integers, but the Fourier series is only comprised of the odd integers.

2. Oct 15, 2008

### buffordboy23

I think I figured it out, or at least obtained a really close approximation.

Multiplying 1/4 to both sides of

$$\sum_{1,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{8}$$

will yield the sum for the terms 2, 6, 10, ...

$$\sum_{2,6,10,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{32}$$

Now multiplying this sum by 1/4 gives the terms 4, 8, 12,...

$$\sum_{4,8,12,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{128}$$

Combining all of the sums together gives a close approximation:

$$\sum_{n}\frac{1}{n^{2}} = \frac{21\pi^{2}}{128} \approx \frac{\pi^{2}}{6}$$

3. Oct 15, 2008

### Dick

If S is the sum over all n (even or odd) then (1/4)S is the sum over all evens. S-(1/4)S is then the sum over all odds which is pi^2/8. Solve for S.

4. Oct 17, 2008

### buffordboy23

EDIT: I see now. Duh! Thanks.

Last edited: Oct 17, 2008