# Series, it's a riddle to me !

1. Nov 26, 2008

### joris_pixie

Hey, guys !
I have to calculate:

$$\sum 1/(2n + 1)^2$$
n from 0 to infinite

Knowing that 1/n² = (pi)²/6

I don't know why I'm stuck here but I just don't get it...

Greetings

Last edited: Nov 26, 2008
2. Nov 26, 2008

### Office_Shredder

Staff Emeritus
Once you have what

$$\sum_{n=1}^{ \infty } \frac{1}{n^2}$$ is, you should be able to find what

$$\sum_{n=1}^{ \infty } \frac{1}{(2n)^2}$$ is pretty easily. And then use both of these to find the series sum that you're looking for

3. Nov 26, 2008

### joris_pixie

It is given ...
Calculate:
$$\sum^{\infty}_{0} 1/(2n +1)^2$$

if you know that:

$$\sum^{\infty}_{1} 1/n^2 = (pi)^2 /6$$

But I can't make it work :(

4. Nov 26, 2008

### Office_Shredder

Staff Emeritus
Can you calculate this?

$$\sum_{n=1}^{ \infty } \frac{1}{(2n)^2}$$

Then what is

$$\sum_{n=1}^{ \infty } \frac{1}{(2n)^2} + \sum_{n=0}^{ \infty } \frac{1}{(2n+1)^2}$$

5. Nov 26, 2008

### joris_pixie

haha I'm sorry for wasting your time !
It was infact not hard, but had a few drinks too much last night.

thank you anyway !