Problems with proof by induction $\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$? 
$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$

Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$
Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Demonstration: $$\frac1{n+1} + \frac1{(n+1)(n+2)}=\dots=\frac1{(n+1)+1}$$
My problem is that I can't find the correct algebra steps to get from the beginning of the demonstration to the end of the demonstration. 

Actually, this is incorrect. The correct answer should be $1-\frac1{n+1}$. 
We can show that the above is true for $n=1$ easily. Now let us show $1-\frac1{n+1}+\frac1{(n+1)(n+2)}=1-\frac1{n+2}$
We can prove that $\frac1{n+1}-\frac1{n+2}=\frac {n+2}{(n+1)(n+2)}-\frac {n+1}{(n+1)(n+2)}=\frac{n+2-n-1}{(n+1)(n+2)}=\frac1{(n+1)(n+2)}$ and vice versa. The $\frac1{n+1}$ terms cancel out, giving us $1-\frac1{n+2}$
