#### Answer

See explanations.

#### Work Step by Step

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 $n$.