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Question Number 68589 by peter frank last updated on 13/Sep/19

Answered by $@ty@m123 last updated on 14/Sep/19

Σ_(k=2) ^n   (1/(n(n+1)))  =Σ_(k=2) ^n  ( (1/n)−(1/(n+1)))  =(1/2)−(1/3)+(1/3)−(1/4)+...+(1/(n−1))−(1/n)+(1/n)−(1/(n+1))  =(1/2)−(1/(n+1))  =((n−1)/(2(n+1)))

$$\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\left(\:\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$=\frac{{n}−\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$

Commented by peter frank last updated on 14/Sep/19

thank you

$${thank}\:{you} \\ $$

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