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

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

Notice that


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


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

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

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