# Facebook's question regarding a sum

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

Prove that

$$\displaystyle \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove that

$$\displaystyle \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}$$

1- By inspection through partial sums

We know that the sum can be written as

$$\displaystyle \sum^{n}_{k=1}\frac{1}{k(k+1)}$$

Let $$\displaystyle S_{n}=\sum^{n}_{k=1}\frac{1}{k(k+1)}$$

$$\displaystyle S_1 = \frac{1}{2}$$

$$\displaystyle S_2=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$$

$$\displaystyle S_3=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{3}{4}$$

$$\displaystyle S_4=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{4}{5}$$

$$\displaystyle S_n=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} \, \cdot\cdot \cdot \,+ \frac{1}{n(n+1)}=\frac{n}{n+1}$$

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2- The second approach is using telescoping series

$$\displaystyle \sum^{n}_{k=1}\frac{1}{k(k+1)}=\sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}$$

$$\displaystyle \sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}=\left( 1 - \frac{1}{2} \right) +\left( \frac{1}{2} - \frac{1}{3} \right) +\left( \frac{1}{3} - \frac{1}{4} \right)+\cdot \cdot \cdot + \left(\frac{1}{n} - \frac{1}{n+1} \right)$$

Cancelling out the terms we reach to

$$\displaystyle \sum^{n}_{k=1}\frac{1}{k}-\frac{1}{k+1}= 1 -\frac{1}{n+1} =\frac{n}{n+1}$$