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Question-163430




Question Number 163430 by amin96 last updated on 06/Jan/22
Answered by qaz last updated on 07/Jan/22
(A/B)=((Σ_(n=1) ^(999) (1/((2n−1)(2n))))/(Σ_(n=1) ^(999) (1/((999+n)(1999−n)))))  =((Σ_(n=1) ^(999) ((1/(2n−1))−(1/(2n)))=H_(1998) −(1/2)H_(999) −(1/2)H_(999) =H_(1998) −H_(999) )/((1/(2998))Σ_(n=1) ^(999) ((1/(999+n))+(1/(1999−n)))=(1/(1499))Σ_(n=1000) ^(1998) (1/n)=(1/(1499))(H_(1998) −H_(999) )))  =1499
$$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{2n}\right)}}{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{999}+\mathrm{n}\right)\left(\mathrm{1999}−\mathrm{n}\right)}} \\ $$$$=\frac{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}}\right)=\mathrm{H}_{\mathrm{1998}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{999}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{999}} =\mathrm{H}_{\mathrm{1998}} −\mathrm{H}_{\mathrm{999}} }{\frac{\mathrm{1}}{\mathrm{2998}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{999}+\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{1999}−\mathrm{n}}\right)=\frac{\mathrm{1}}{\mathrm{1499}}\underset{\mathrm{n}=\mathrm{1000}} {\overset{\mathrm{1998}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}=\frac{\mathrm{1}}{\mathrm{1499}}\left(\mathrm{H}_{\mathrm{1998}} −\mathrm{H}_{\mathrm{999}} \right)} \\ $$$$=\mathrm{1499} \\ $$

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