Proving a relation between $\sum\frac{1}{(2n-1)^2}$ and $\sum \frac{1}{n^2}$ I ran into this question:

Prove that:
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Thank you very much in advance.

Hint: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}$$
