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!

$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$ Do you see how to do it now?
