How do I prove by induction that, for $n≥1, \sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}$? Hi can you help me solve this:
I have proved that $p(1)$ is true and am now assuming that $p(k)$ is true. I just don't know how to show $p(k+1)$ for both sides?

HINT:
$$\frac1{r(r+1)}=\frac1r-\frac1{r+1}$$
and
$$\frac{n}{n+1}=1-\frac1{n+1}\;.$$
