Prove the following series

[SilverKing]

Hi everyone,

I've this problem:

 

[pka]

Hi everyone,
I've this problem:

From (2) we get \(\displaystyle \dfrac{{{\pi ^2}}}{{12}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n+1 }}{n^{ - 2}}} \) (4)

(3) is correct \(\displaystyle \dfrac{{{\pi ^2}}}{6} = \sum\limits_{n = 1}^\infty {{n^{ - 2}}} \)

Now ADD (3)+(4): \(\displaystyle \dfrac{{{\pi ^2}}}{4} = \sum\limits_{n = 1}^\infty {{2(2n-1)^{-2}}} \)

Divide by 2.