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Question Number 93859 by  M±th+et+s last updated on 15/May/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n(n+1)}$$

Answered by Ar Brandon last updated on 15/May/20

$$\frac{1}{\mathrm{n}(\mathrm{n}+1)}=\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\Rightarrow \underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}(\mathrm{n}+1)}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}]=\underset{\mathrm{k}\to +\mathrm{\infty }}{\lim }\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}[\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}]$$$$=\underset{\mathrm{k}\to +\mathrm{\infty }}{\lim }[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdot \cdot \cdot +\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1}]$$$$=\underset{\mathrm{k}\to +\mathrm{\infty }}{\lim }[1-\frac{1}{\mathrm{k}+1}]=1$$

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