# Thread: Limit of this series?

1. ## Limit of this series?

$\displaystyle \sum_{n=0}^{\infty}(\frac{x}{2})^{2n+1}$

Not sure how to go abou this..

Any help?

2. This series looks like a geometric series...

$\displaystyle \sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n+1} =\frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{4} \right)^{n}$
Does it help ?

3. Aha! Thank you, I got it

4. How did you do it?

It looks like it might involve cosh (does it?).

5. Originally Posted by Showcase_22
How did you do it?
$\displaystyle \sum_{n=0}^{\infty}\left(\frac{x^2}{4}\right)^{n}$ is a geometric series (common ratio : $\displaystyle x^2/4$) hence

$\displaystyle \frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{4}\ right)^{n}=\frac{x}{2}\cdot \frac{1}{1-\frac{x^2}{4}}=\frac{2x}{4-x^2}\quad \text{\ if\ } \left|\frac{x^2}{4}\right|<1$

6. ahh, i'm clearly getting a bit rusty after my exam.

Thanks!

7. Originally Posted by Showcase_22
How did you do it?

It looks like it might involve cosh (does it?).
It would, if there were $\displaystyle (2n)!$ in the denominator