How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} +  \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$?

Write
$$\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1},$$
split the sum into two sums, shift the index of one, recombine, simplify, and you are done. 
N.B. You need to do this for partial sums and only at the end take the limit to be rigorous.
