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?

$$\sum_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$$
for $n=1$ we have $\frac{1}{1(1+1)}=\frac{1}{1+1}$ suppose that
$$\sum_{r=1}^{k}\frac{1}{r(r+1)}=\frac{k}{k+1}$$ then
$$\sum_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)(k+2)}=$$
$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=$$
$$=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{(k+1)}{(k+1)+1}$$ 
