Summation

• Nov 21st 2010, 11:00 AM
dwsmith
Summation
Use the identity $\displaystyle \frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}$ to derive a formula for $\displaystyle \sum_{i=1}^{n}\frac{1}{i(i+1)}$.

I am not sure how to do this.
• Nov 21st 2010, 11:02 AM
Also sprach Zarathustra
Develop the sum for n=4... what do you see?
• Nov 21st 2010, 11:12 AM
Also sprach Zarathustra
$\displaystyle \sum_{i=1}^{n}\frac{1}{i(i+1)}=\displaystyle \sum_{i=1}^{n}\frac{1}{i}-\frac{1}{i+1}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n-1}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{n+1}$

By the way it's very famous series!

Telescoping series - Wikipedia, the free encyclopedia