# Formula of a summation

• Apr 19th 2009, 08:42 PM
vexiked
Formula of a summation
Hi all... I am having alot of trouble finding a guide on how to solve this. I have been told that I can compute the first few terms to get the formula but have no idea where to start. Thanks for your help in advance!
Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1).
• Apr 19th 2009, 10:14 PM
Chris L T521
Quote:

Originally Posted by vexiked
Hi all... I am having alot of trouble finding a guide on how to solve this. I have been told that I can compute the first few terms to get the formula but have no idea where to start. Thanks for your help in advance!
Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1).

In summation form, this is $\displaystyle \sum_{k=1}^n\frac{1}{k(k+1)}$

If you do a partial fraction decomposition on $\displaystyle \frac{1}{k(k+1)}$, you should end up with $\displaystyle \frac{1}{k}-\frac{1}{k+1}$

Then, $\displaystyle \sum_{k=1}^n\frac{1}{k(k+1)}=\sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]$

When you expand the sum, you end up with $\displaystyle \sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\frac{1}{n}-\frac{1}{n+1}$. Now, a lot of cancellations occur, and we are left with $\displaystyle 1-\frac{1}{n+1}$

So we can say that $\displaystyle \sum_{k=1}^n\frac{1}{k(k+1)}=\sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{n+1}$

Does this make sense?
• Apr 19th 2009, 10:19 PM
chisigma
Since is...

$\displaystyle \frac{1}{k\cdot (k+1)}= \frac{1}{k}-\frac{1}{k+1}$

... il also...

$\displaystyle \sum_{k=1}^{n} \frac{1}{k\cdot (k+1)}= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{n} - \frac{1}{n+1} = 1-\frac{1}{n+1}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$