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

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.

## marked as duplicate by Martin Sleziak, Community♦Jan 7 '17 at 19:57

• Start with partial fraction decomposition. – Simply Beautiful Art Nov 17 '16 at 22:07
• In other words, $\frac1{n(n+1)} = \frac{?}n + \frac{?}{n+1}$. – Théophile Nov 17 '16 at 22:08
• Since you are a new user to the site... WELCOME! And to avoid any more $\LaTeX$ issues, check the handbook. – Simply Beautiful Art Nov 17 '16 at 22:26
• Without induction? Well, isn't that my answer below? – Simply Beautiful Art Nov 17 '16 at 23:11
• yes it is I just didnt have time to select the answer. Thank you very much for your help! very much appreciated. – Hidaw Nov 17 '16 at 23:14

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.

By induction,

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

then

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

• @Flow: it does not obviously. It proves an answer that you knew. – Yves Daoust Nov 17 '16 at 22:50