Finding the formula of the sum $\frac1{1\cdot2} + \frac1{2\cdot3}+ \frac1{3\cdot4} + \cdots + \frac1{n\cdot(n+1)}$?

Question

I am having some trouble finding out the formula for this sum:

$$\text{?} = \frac1{1\cdot2} + \frac1{2\cdot3}+ \frac1{3\cdot4} + \cdots + \frac1{n\cdot(n+1)}$$

I am not sure where to start finding the formula. I know the answer is $1/(n+1)$ but how do you get that without using INDUCTION.

Answer 1

Notice that

$$\frac1{n(n+1)}=\frac1n-\frac1{n+1}$$

This makes this a telescoping sum:

$$\begin{align}S&=\quad\frac1{1\times2}\ \ \ \quad+\frac1{2\times3}\ \ \ \ \ \ \ \ +\frac1{3\times4}\ \ \ +\dots+\quad\ \frac1{n(n+1)}\\&=\left(\frac11-\color{#ee8844}{\frac12}\right)+\left(\color{#ee8844}{\frac12}-\color{#559999}{\frac13}\right)+\left(\color{#559999}{\frac13}-\color{#034da3}{\frac14}\right)+\dots+\left(\color{#034da3}{\frac1n}-\frac1{n+1}\right)\\&=1-\frac1{n+1}\end{align}$$

Since each colored term cancels with the next.

Answer 2

By induction,

If $$S_n=\frac n{n+1}$$

then

$$S_{n+1}=S_n+\frac1{(n+1)(n+2)}=\frac n{n+1}+\frac1{(n+1)(n+2)}=\frac{n+1}{n+2}.$$