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!
 A: $$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$ Do you see how to do it now?
A: Since you speak of the ratio test, maybe you're only concerned with whether the series converges rather than with what the sum is.  That is the most that the ratio test can give you.  If that is what you're concerned with, then you can say
$$
\sum_{k=1}^\infty \frac 1 {k(k+1)} \le \sum_{k=1}^\infty \frac 1 {k^2}
$$
and that converges, by a simple integral test.
However, we can also say
$$
\frac 1 {k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
$$
and then we have
\begin{align}
& \overbrace{\left(\frac 1 1 - \frac 1 2\right)}^{k=1} + \overbrace{\left(\frac 1 2 - \frac 1 3\right)}^{k=2} + \overbrace{\left(\frac 1 3 - \frac 1 4\right)}^{k=3} + \cdots + \overbrace{\left( \frac 1 n - \frac 1 {n+1} \right)}^{k=n} \\[10pt]
= {} & \frac 1 1 \underbrace{{} - \frac 1 2 + \frac 1 2 }\ \ \underbrace{{}-\frac 1 3 + \frac 1 3 }\ \ \underbrace{{} - \frac 1 4 + \frac 1 4} -\cdots\cdots\underbrace{{} - \frac 1 n + \frac 1 n} - \frac 1 {n+1} \\[10pt]
= {} & 1 - \frac 1 {n+1} \to 1 \text{ as }n\to\infty.
\end{align}
A: $$\sum_{k=1}^n\frac{1}{k(k+1)} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=\frac{n}{n+1}$$
So 
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k(k+1)}= \lim_{n\to \infty} \frac{n}{n+1}=1.$$
