Sum as formula

1. Oct 21, 2011

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.

2. Oct 21, 2011

Mute

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?

3. Oct 21, 2011

lurflurf

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

4. Oct 22, 2011

HallsofIvy

Staff Emeritus
If I understand what you are asking, it is just
$$\sum_{i= 1}^n\frac{1}{i}+ \sum_{i= 1}^n\frac{1}{i+ 1}= \sum_{i= 1}^n\left(\frac{1}{i}- \frac{1}{i+ 1}\right)$$
$$= \sum_{i= 1}^n \left(\frac{i+1}{i(i+1)}- \frac{i}{i(i+1)}\right)= \sum_{i= 1}^n\frac{1}{i(i+1)}$$

In other words, combine the two sums, get common denominators, and subtract the numerators.

5. Oct 22, 2011

mathman

Telescoping result. 1 - 1/(n+1) = n/(n+1)