Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)))))

limx(1+12+13+......+1n1+13+15......+12n+1)

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

1+12+13+....+1n=Hn1+13+15+....+12n+1=1+12+13+.....+12n+12n+11214...12n=H2n+112Hnqn=HnH2n+1Hn2=1H2n+1Hn12wehaveH2n+1Hn=ln(2n+1)+γ+o(12n+1)ln(n)+γ+o(1n)=ln(2n+1)ln(n)×1+γln(2n+1)+o(1(2n+1)ln(2n+1))1+γln(n)+o(1nln(n))limn+H2n+1Hn=limn+ln(2n+1)ln(n)=limn+ln(n)+ln(2+1n)ln(n)=limn+1+ln(2+1n)ln(n)1=1limn+qn=1112=2

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

well done

welldone

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

you are welcome

youarewelcome

Terms of Service

Privacy Policy

Contact: info@tinkutara.com