Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$ 
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$ 

I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.
Please help___.

Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .
$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$ Hope it helps!!!
