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Question Number 112881 by bemath last updated on 10/Sep/20

Answered by Dwaipayan Shikari last updated on 10/Sep/20

(1/1)+(1/(1+2))+(1/(1+2+3))+(1/(1+2+3+4))...  Σ_(n=1) ^n (2/(n(n+1)))=2Σ_(n=1) ^(40) (1/n)−(1/(n+1))=2(1−(1/(41)))=((80)/(41))

$$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}... \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{40}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{41}}\right)=\frac{\mathrm{80}}{\mathrm{41}} \\ $$

Commented by bemath last updated on 10/Sep/20

sir how you to get the formula (2/(n(n+1)))

$$\mathrm{sir}\:\mathrm{how}\:\mathrm{you}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{formula}\:\frac{\mathrm{2}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$

Commented by Dwaipayan Shikari last updated on 10/Sep/20

Term nth  =(1/(1+2+3+4+5+6+7+...+n))=(1/((n(n+1))/2))=(2/(n(n+1)))

$${Term}\:{nth} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+...+{n}}=\frac{\mathrm{1}}{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}}=\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$$ \\ $$

Commented by bemath last updated on 10/Sep/20

oo thank you

$$\mathrm{oo}\:\mathrm{thank}\:\mathrm{you} \\ $$

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