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Question Number 101239 by  M±th+et+s last updated on 01/Jul/20

lim_(x→∞) (((1+(1/2)+(1/3)+......+(1/n))/(1+(1/3)+(1/5)......+(1/(2n+1)))))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+......+\frac{\mathrm{1}}{{n}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}......+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}}\right) \\ $$

Answered by mathmax by abdo last updated on 01/Jul/20

1+(1/2)+(1/3)+....+(1/n) =H_n 1+(1/3)+(1/5)+....+(1/(2n+1)) =1+(1/2)+(1/3)+.....+(1/(2n))+(1/(2n+1)) −(1/2)−(1/4)−...−(1/(2n)) =H_(2n+1) −(1/2)H_n ⇒q_n =(H_n /(H_(2n+1) −(H_n /2))) =(1/((H_(2n+1) /H_n )−(1/2))) we have (H_(2n+1) /H_n ) =((ln(2n+1)+γ +o((1/(2n+1))))/(ln(n)+γ +o((1/n)))) =((ln(2n+1))/(ln(n)))×((1+(γ/(ln(2n+1)))+o((1/((2n+1)ln(2n+1)))))/(1+(γ/(ln(n)))+o((1/(nln(n)))))) ⇒ lim_(n→+∞) (H_(2n+1) /H_n ) =lim_(n→+∞) ((ln(2n+1))/(ln(n))) =lim_(n→+∞) ((ln(n)+ln(2+(1/n)))/(ln(n))) =lim_(n→+∞) ((1+((ln(2+(1/n)))/(ln(n))))/1) =1 ⇒lim_(n→+∞) q_n =(1/(1−(1/2))) =2

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+....+\frac{\mathrm{1}}{\mathrm{n}}\:=\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+....+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+.....+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−...−\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$=\mathrm{H}_{\mathrm{2n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{n}} \:\Rightarrow\mathrm{q}_{\mathrm{n}} =\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{H}_{\mathrm{2n}+\mathrm{1}} −\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{2}}}\:=\frac{\mathrm{1}}{\frac{\mathrm{H}_{\mathrm{2n}+\mathrm{1}} }{\mathrm{H}_{\mathrm{n}} }−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\frac{\mathrm{H}_{\mathrm{2n}+\mathrm{1}} }{\mathrm{H}_{\mathrm{n}} }\:=\frac{\mathrm{ln}\left(\mathrm{2n}+\mathrm{1}\right)+\gamma\:+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right)}{\mathrm{ln}\left(\mathrm{n}\right)+\gamma\:+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)} \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{ln}\left(\mathrm{n}\right)}×\frac{\mathrm{1}+\frac{\gamma}{\mathrm{ln}\left(\mathrm{2n}+\mathrm{1}\right)}+\mathrm{o}\left(\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2n}+\mathrm{1}\right)}\right)}{\mathrm{1}+\frac{\gamma}{\mathrm{ln}\left(\mathrm{n}\right)}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{nln}\left(\mathrm{n}\right)}\right)}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{H}_{\mathrm{2n}+\mathrm{1}} }{\mathrm{H}_{\mathrm{n}} }\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{ln}\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{ln}\left(\mathrm{n}\right)}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{ln}\left(\mathrm{n}\right)+\mathrm{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{n}}\right)}{\mathrm{ln}\left(\mathrm{n}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\frac{\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{n}}\right)}{\mathrm{ln}\left(\mathrm{n}\right)}}{\mathrm{1}}\:=\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{q}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{2} \\ $$

Commented by  M±th+et+s last updated on 01/Jul/20

well done

$${well}\:{done} \\ $$

Commented by mathmax by abdo last updated on 01/Jul/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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