Proving Summation of \frac{1}{i(i+1)} = \frac{n}{n+1}

[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.

 

[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?

 

[lurflurf]

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

 

[HallsofIvy]

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.

 

[mathman]

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