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)))))

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

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

well done

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

you are welcome