# Value of a inf. series

1. Dec 6, 2005

### standardflop

Hi,

by calculating the fourier series for |sin x| it was found that,
$$\sum_{n=1}^\infty \frac 1{(2n-1)(2n+1)}=1/2$$.
How can i find the value of -
$$\sum_{n=1}^\infty \frac 1{(2n-1)^2(2n+1)^2}$$ ?

Thanks

2. Dec 6, 2005

### StatusX

I assume you found the value of the first sum by expanding each term to get a series where almost every term cancels. Use the same expansion, but now you need to square it. Expanding the square will give you three terms, one of which is just the original series, and the other two involve sums of the odd inverse squares, ie, 1+1/3^2+1/5^2+... . These can be found by noting that the sum of all inverse squares is pi^2/6, and you can take a common factor out of the even squares to get their value, and subtract this to get the sum you want.

3. Dec 6, 2005

### standardflop

No, actually, i found that

$$|\sin x| = \frac 2{\pi} - \frac 4{\pi} \sum_{n=1}^\infty \frac{\cos (2nx)}{(2n+1)(2n-1)}, \forall x \in \mathbb{R}$$
and used this for x=0. But you cant "just" sqare under the 'sum', right?

4. Dec 7, 2005

### StatusX

The expansion I thought you used was:

$$\frac{1}{(2n+1)(2n-1)} = \frac{1}{2} \left( \frac{1}{2n-1}-\frac{1}{2n+1} \right)$$

This cancels at every term except the first. By squaring this, you get a term that looks like the original series and two other terms involving odd squares.

5. Dec 7, 2005

### siddharth

$$\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)$$

6. Dec 7, 2005

### standardflop

Alright, then if i got it right, you mean to say that

$$\sum_{n=1}^\infty \frac 1{(2n-1)^2(2n+1)^2} = -2 \sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)} + \sum_{n=1}^\infty \frac 1{(2n-1)^2}+\sum_{n=1}^\infty \frac 1{(2n+1)^2}$$
How how do you evaluate the two inverse squared sums... seems like i just got two ekstra problems? :)

7. Dec 7, 2005

### StatusX

Right. Now use the fact that:

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

Find the even sum by taking out a common factor and then subtract this from the above value.