# 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!

## 2 Answers

By direct calculation, we see that \begin{align} \sum^\infty_{n=1} \frac{1}{n^2} = \sum^\infty_{k=1}\frac{1}{(2k-1)^2}+\frac{1}{4}\sum^\infty_{n=1}\frac{1}{n^2} \end{align} which means \begin{align} \sum^\infty_{k=1} \frac{1}{(2k-1)^2}=\frac{3}{4}\sum^\infty_{n=1}\frac{1}{n^2} = \frac{3}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}. \end{align}

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.