Series, it's a riddle to me !

[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

The answer is (pi)²/8

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

Greetings

[Office_Shredder]

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

[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 :(

[Office_Shredder]

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}

[joris_pixie]

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

http://www.pixie.be/solution.JPG


thank you anyway !