# Thread: Telescoping series

1. ## Telescoping series

Question:
If the following series converges, compute its sum. Otherwise, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, and DIV otherwise.

I know that I can get 1(2n+4) - 1/2n as the partial fraction decomposition within the sum and that I need to write the terms to see the pattern and all that so that's not my problem. My problem is that I'm having trouble seeing the pattern and would appreciate it if someone could show me their work and explain their logic.

Any help would GREATLY appreciated!
Thanks in advance!

2. Originally Posted by s3a
Question:

If the following series converges, compute its sum. Otherwise, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, and DIV otherwise.

I know that I can get 1(2n+4) - 1/2n as the partial fraction decomposition within the sum and that I need to write the terms to see the pattern and all that so that's not my problem. My problem is that I'm having trouble seeing the pattern and would appreciate it if someone could show me their work and explain their logic.

Any help would GREATLY appreciated!
Thanks in advance!

$\sum^\infty_{n=1}\frac{1}{n(n+2)}=\sum^\infty_{n=1 }\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$ $=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots\right]$.

Well, now cancel whatever possible and check the limit ...

Tonio

3. $\frac{1}{n(n+2)}=\frac{1}{2}\left( \frac{n+2-n}{n(n+2)} \right)=\frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2} \right).$

4. Originally Posted by tonio
$\sum^\infty_{n=1}\frac{1}{n(n+2)}=\sum^\infty_{n=1 }\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$ $=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots\right]$.

Well, now cancel whatever possible and check the limit ...

Tonio
Do I calculate the limit of 1/2 + 1/4 - 1/(2n+4) as n->inf ? (I know this answer is numerically correct but given that both parts of the sum go to zero, I'd just like to know if I chose the correct one.)