# Thread: Series ( tricky equation )

1. ## Series ( tricky equation )

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

2. ## Re: Series ( tricky equation )

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 \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*}

3. ## Re: Series ( tricky equation )

Thanks , this makes perfect sense!