Find the limit: lim_{nrightarrowinfty}(frac{1}{n+1}+frac{1}{n+2}+...+frac{1}{2n})

Trent Carpenter 2020-10-27 Answered
Find the limit:
limn(1n+1+1n+2+...+12n)
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Expert Answer

oppturf
Answered 2020-10-28 Author has 94 answers
Consider the given limit,
limn(1n+1+1n+2+...+12n)
now write the limit in the summation form,
limn(1n+1+1n+2+...+12n)=limnr=1n(1n+r)
=limnr=1n1n(11+rn)
now, convert it into definite integral as follows,
substitute rn=xdx=1n
so the limits of integration will be
r1,x0
rn,x1
so,
limnr=1n1n(11+rn)=0111+xdx
=[ln(1+x)]01
=(ln(2)ln(1))
=ln(2)
hence the value of the limit will be ln(2).
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Jeffrey Jordon
Answered 2022-06-04 Author has 2070 answers

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