# Homework Help: Fourier Series - proving a sum

1. Jun 16, 2010

### ramdayal9

1. The problem statement, all variables and given/known data
Let $f(x)=x$ on $[-\pi,\pi)$ and peridically extended. Compute the fourier series and hence show:

$\sum_{n \geq 1,nodd} \frac{1}{n^2} = \frac{\pi^2}{8}$ and $\sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6}$

2. Relevant equations

Parseval's equality

3. The attempt at a solution
I computed the fourier series to be $-\sum_{n=1,nodd} \frac{4}{n^2 \pi} e^{inx}+\frac{\pi}{2}$ (even terms $\hat{f}(n)=0$) and proved the first sum (letting x=0).
How would I compute the second part? how do i get the whole sum from this? I tried to slipt the sum into even and odd parts, but i dont know how to compute the even sum when I dont have any terms for the even sum! thanks

2. Jun 16, 2010

### HallsofIvy

You can't. If, in fact, all terms with even indices are 0, then "$\sum_{n\ge 1, n odd}$" and $\sum_{n\ge 1}$ must be the same!

3. Jun 16, 2010

### ramdayal9

Ok, so how would I use what I have to prove the second sum? That's assuming that fourier series is right - I think it is, because the first sum works out. On a related note, how would I go about proving that $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} = \frac{\pi^4}{96}$? I have started by saying that this is equivalent to the sum $\sum_{n \geq 1, n odd} \frac{1}{n^4}$. I use a similar method as before and consider the function $f(x) = |x^3|$ on $[-\pi, \pi)$ but I dont seem to get the sum I require in the end...I get the sum of even numbers instead!
Thanks

4. Jun 16, 2010

### vela

Staff Emeritus
I don't get the same Fourier series for f(x)=x that you do.