Telescoping Series

One type of series that lends itself to easy formulation of $S_n$ are called telescoping series. Here is an example using partial fractions.

$$\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n(n+1)} = \displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n}-\displaystyle \frac{1}{n+1}$$

$$= \left(\displaystyle \frac{1}{1}-\displaystyle \frac{1}{2}\right)+... ...\right)+ \left(\displaystyle \frac{1}{3}-\displaystyle \frac{1}{4}\right)+\dots$$

As we add more and more terms to the series everythig cancels out except the first and last terms. So we find

$$S_n=1-\displaystyle \frac{1}{n+1}$$

After adding up 100 terms we will have reached $S_{100}=1-1/101$. It follows from what we have already said that

$$\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{1}{n(n+1)}=\displaystyle \lim_{n\to\infty}S_n=1$$