How to solve a particular well ordering problem I want to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$ with well ordering principle. 
I know that $$n \in N $$
because the numbers 1,2,3... are discrete and are positive. I was thinking of using a set of natural numbers that don't satisfy the equation and proving by contradiction  but I'm not sure that works. 

Yes, you can make it like that. It is not different than induction, though. 
Let $E$ be the set of $n\in\mathbb N$ such that your equality does not hold. Note that $1\not\in E$. Let $k\geq2$ be the minimum of $E$ (here is where one uses the principle). Then $k-1\not\in E$, so it satisfies the formula:
$$
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}=\frac{k-1}{k}.
$$
But then
\begin{align}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{k(k+1)}&=
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}+ \frac{1}{k(k+1)}\\ \ \\
&=\frac{k-1}k+\frac{1}{k(k+1)}=\frac{(k-1)(k+1)+1}{k(k+1)}\\ \ \\
&=\frac{k^2}{k(k+1)}=\frac{k}{k+1}.
\end{align}
Then $k$ would satisfy the equality a contradiction. That is, the minimum of $E$ cannot exist. So it has to be that $E$ is empty, and the formula holds for all $n\in\mathbb N$. 
