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Question Number 101239 by M±th+et+s last updated on 01/Jul/20
limx→∞(1+12+13+......+1n1+13+15......+12n+1)
Answered by mathmax by abdo last updated on 01/Jul/20
1+12+13+....+1n=Hn1+13+15+....+12n+1=1+12+13+.....+12n+12n+1−12−14−...−12n=H2n+1−12Hn⇒qn=HnH2n+1−Hn2=1H2n+1Hn−12wehaveH2n+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=1⇒limn→+∞qn=11−12=2
Commented by M±th+et+s last updated on 01/Jul/20
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Commented by mathmax by abdo last updated on 01/Jul/20
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