# Computing a series with another series

I was asked to compute this series:

$\sum _{n=1}^{\infty }\:\frac{1}{\left(2n-1\right)^2}$

but by using the fact that $\sum _{n=1}^{\infty }\:\frac{1}{n^2}=\frac{\pi ^2}{6}$

I think i know how to compute the series by itself by using the telescoping test but I am not sure how to use the second series...I know it must have something to do with power series but any help to get started would be great!

Observe that $$\sum_{n=1}^\infty\frac{1}{(2n)^2}=\frac{1}{(2\cdot1)^2}+\frac{1}{(2\cdot2)^2}+\frac{1}{(2\cdot3)^2}\dotsb =\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dotsb\right)=\frac{\pi^2}{4(6)}.$$ But $$\sum_{n=1}^\infty\frac{1}{(2n)^2}+\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\sum_{n=1}^\infty\frac{1}{n^2}$$ where rearrangement is allowed since we are dealing with non-negative series.