1. ## simple sigma problem

No. 20 / Thank you for helping me

2. Originally Posted by narbe

No. 20 / Thank you for helping me
Note that $\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^n \left[\frac{1}{k}-\frac{1}{k-1}\right]$ is a telescoping series. So when you go to expand it, we are left with:

$\underbrace{1-\frac{1}{2}}_{k=1}+\underbrace{\frac{1}{2}-\frac{1}{3}}_{k=2}+\underbrace{\frac{1}{3}-\frac{1}{4}}_{k=3}+\dots+\underbrace{\frac{1}{n}-\frac{1}{n+1}}_{k=n}$

So we see that all the terms cancel out, except for the first and last terms.

So $\sum_{k=1}^n \left[\frac{1}{k}-\frac{1}{k-1}\right]=\color{red}\boxed{1-\frac{1}{n+1}}$

Does this make sense?

--Chris