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?
 A: 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.
A: If you want to do it by a conventional "blind" induction, suppose that for a certain $k$ we have
$$\sum_1^k \frac{1}{r(r+1)}=\frac{k}{k+1}.\tag{$1$}$$
We want to prove that
$$\sum_1^{k+1} \frac{1}{r(r+1)}=\frac{k+1}{k+2}.\tag{$2$}$$
Note that the left-hand side of $(2)$ is the left-hand side of $(1)$, plus $\dfrac{1}{(k+1)(k+2)}$.
So we want to prove that 
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}.\tag{$3$}$$
It seems reasonable to manipulate the left-hand side of $(3)$ and see whether we get the right-hand side. A common manipulation is to bring the expression to the common denominator $(k+1)(k+2)$. We get
$$\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}.$$
This is equal to $\dfrac{k^2+2k+1}{(k+1)(k+2)}$.
But the numerator is equal to $(k+1)^2$. Cancel a $k+1$.   
Remark: The algebra at the end is neater, and closer to the informal "telescoping" argument, if we observe that $\dfrac{1}{(k+1)(k+2)}=\dfrac{1}{k+1}-\dfrac{1}{k+2}$. Thus
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+ \frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}=\frac{k+1}{k+2}.$$
A: $$\sum_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$$
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
$$\sum_{r=1}^{k+1}\frac{1}{r(r+1)}=\sum_{r=1}^{k}\frac{1}{r(r+1)}+\frac{1}{(k+1)(k+2)}=$$
$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=$$
$$=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{(k+1)}{(k+1)+1}$$ 
A: HINT:
$$\frac1{r(r+1)}=\frac1r-\frac1{r+1}$$
and
$$\frac{n}{n+1}=1-\frac1{n+1}\;.$$
