find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!

$$\sum_{k=1}^n\frac{1}{k(k+1)} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=\frac{n}{n+1}$$
So 
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k(k+1)}= \lim_{n\to \infty} \frac{n}{n+1}=1.$$
