Question: 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?

Answer: 
Hint: $$\sum\_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum\_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)((k+1)+1)}.$$ Now use the inductive hypothesis and see if that gets you anywhere.
$$\sum\_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$$
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
$$\sum\_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum\_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)(k+2)}=$$
$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=$$
$$=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{(k+1)}{(k+1)+1}$$