Use Mathematical Induction to prove that $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$ [duplicate]

Let $\displaystyle f(n):\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{n(n+1)}=1-\frac1{n+1}$

holds true for $n=m$

$\displaystyle\implies f(m):\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{m(m+1)}=1-\frac1{m+1}\ \ \ \ (1)$

For $n=m+1$ adding $\displaystyle\frac1{(m+1)(m+2)}$ to both sides of $(1),$ $\displaystyle\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{m(m+1)}+\frac1{(m+1)(m+2)}=1-\frac1{m+1}+\frac1{(m+1)(m+2)}$

Now $\displaystyle\frac1{(m+1)(m+2)}=\frac{(m+2)-(m+1)}{(m+1)(m+2)}=\frac1{m+1}-\frac1{m+2}$

or $\displaystyle-\frac1{m+1}+\frac1{(m+1)(m+2)}=\frac{-(m+2)+1}{(m+1)(m+2)}=\cdots$

Can you take it home from here?

Don't forget to demonstrate the base case