Consider the following series

$$S = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + \cdots$$

we can write this compactly as 

$$S = \sum_{n =1}^{\infty} \frac{1}{n(n + 1)}$$

decomposing into partial fractions give

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

written out to the $n$ th term this series is

$$S_n = 1\mathbin{-} \frac{1}{2} + \frac{1}{2}\mathbin{-}\frac{1}{3} + \frac{1}{3}\mathbin{-}\frac{1}{4} + \frac{1}{4}\mathbin{-} \cdots + \frac{1}{n}\mathbin{-} \frac{1}{1 + n}$$

now on the right-hand side all the middle terms cancel out and the expression collapses to a simple form like a pocket telescope that can be extended and collapsed to a small size after use.

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

$$S = \lim_{n\to\infty}\left(1\mathbin{-}\frac{1}{n + 1}\right) = 1$$

Exercises:

Sum the following series

$$\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\cdots$$

$$\frac{1}{1\times2}\mathbin{-}\frac{3}{2\times3}+\frac{7}{3\times4}\mathbin{-}\frac{9}{4\times5}+\cdots$$