Question Number 93859 by M±th+et+s last updated on 15/May/20
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$
Answered by Ar Brandon last updated on 15/May/20
![(1/(n(n+1)))=(1/n)−(1/(n+1))⇒Σ_(n=1) ^∞ (1/(n(n+1)))=Σ_(n=1) ^∞ [(1/n)−(1/(n+1))]=lim_(k→+∞) Σ_(n=1) ^k [(1/n)−(1/(n+1))] =lim_(k→+∞) [1−(1/2)+(1/2)−(1/3)+(1/3)−(1/4)+∙∙∙+(1/k)−(1/(k+1))] =lim_(k→+∞) [1−(1/(k+1))]=1](https://www.tinkutara.com/question/Q93861.png)
$$\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\Rightarrow\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right]=\underset{\mathrm{k}\rightarrow+\infty} {\mathrm{lim}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right] \\ $$$$=\underset{\mathrm{k}\rightarrow+\infty} {\mathrm{lim}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right] \\ $$$$=\underset{\mathrm{k}\rightarrow+\infty} {\mathrm{lim}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right]=\mathrm{1} \\ $$