# Express as a telescoping series. If convergent find its sum.

• Nov 18th 2012, 09:11 PM
superduper1
Express as a telescoping series. If convergent find its sum.
$\sum\limits_{n=1}^{\infty} \frac{1}{n(n+2)}$

I used partial fractions and got this term:

$\sum\limits_{n=1}^{\infty} \frac{1}{n(n+2)} = \frac{\frac{1}{2}}{n} - \frac{\frac{1}{2}}{n+2}$

Then factored out the 1/2 and I'm began to expand,

$\frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+2} = ....$

After taking the limit as n goes to infinity I eventually get that the answer converges to 3/4. Can someone please confirm?
• Nov 19th 2012, 01:53 AM
hollywood
Re: Express as a telescoping series. If convergent find its sum.
This is why the series is called "telescoping":

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

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

$\frac{1}{2}\left(\sum\limits_{n=1}^{\infty} \frac{1}{n} - \sum\limits_{n=3}^{\infty}\frac{1}{n} \right)=$

$\frac{1}{2}\left( \frac{1}{1} + \frac{1}{2} \right)$

- Hollywood