<s>[INST] 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?
 [/INST] 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
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+ \frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}=\frac{k+1}{k+2}.$$
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