## Precalculus (6th Edition) Blitzer

Step 1. Check if it is true for $n=1$. We have $\frac{1}{1\cdot2}=\frac{1}{1+1}$ which is true. Step 2. Assume that it is true for $n=k$. We have $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}++\frac{1}{k(k+1)}=\frac{k}{k+1}$ Step 3. For $n=k+1$, we have $LHS=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}++\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{k+1}{k+2}$ and $RHS=\frac{k+1}{k+1+1}=LHS$ Thus, it is also true for $n=k+1$ Step 4. Thus, through mathematical induction, we proved that the formula is true for all integer $n$.