# Sum as formula

neom
How can I show that $$\sum_{i=1}^n\;\frac1{i(i+1)}=\frac{n}{n+1}$$

I've already figured out i can write it as $$\sum_{i=1}^n\;\frac1{i}-\sum_{i=1}^n\;\frac1{i+1}$$

but as I'm a little drunk I can't figure out how to get from there to the formula.

Sorry if I put this in the wrong sextion, but twas in my calculus book.

Homework Helper
1) Don't drink and derive.

2) What would happen, say, if you wrote the first sum as $1 + \sum_{i=2}^n (1/i)$ and then changed variables in that sum to something convenient?

Homework Helper
It is called a telescoping series. The idea is only one term from each sum remain after all others cancel.

$$\sum_{i= 1}^n\frac{1}{i}+ \sum_{i= 1}^n\frac{1}{i+ 1}= \sum_{i= 1}^n\left(\frac{1}{i}- \frac{1}{i+ 1}\right)$$
$$= \sum_{i= 1}^n \left(\frac{i+1}{i(i+1)}- \frac{i}{i(i+1)}\right)= \sum_{i= 1}^n\frac{1}{i(i+1)}$$