Having a lot of issue with proving this series:
Given ∑ 1/n^2 converges to pi^2/6
Show that ∑ 1/ (2n-1)^2 converges to pi^2/8
Both equations have limits n=1 to infinity
Any help would be greatly appreciated
The sum you are trying to evaluate is the sum of reciprocals of squares of all the odd integers.
You can evaluate this by using the sum of reciprocals of squares, and subtracting the sum of reciprocals of squares of the even integers.
$\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{1}{(2n - 1)^2}} &= \sum_{n = 1}^{\infty}{\frac{1}{n^2}} - \sum_{n = 1}^{\infty}{\frac{1}{(2n)^2}} \\ &= \frac{\pi^2}{6} - \sum_{n = 1}^{\infty}{\frac{1}{4n^2}} \\ &= \frac{\pi^2}{6} - \frac{1}{4}\sum_{n = 1}^{\infty}{\frac{1}{n^2}} \\ &= \frac{\pi^2}{6} - \frac{1}{4}\cdot \frac{\pi^2}{6} \\ &= \frac{\pi^2}{6} - \frac{\pi^2}{24} \\ &= \frac{4\pi^2}{24} - \frac{\pi^2}{24} \\ &= \frac{3\pi^2}{24} \\ &= \frac{\pi^2}{8} \end{align*} $