Dear Student

the sum can be written as ∑_r=1^r=n  1/(r + √rn)

= ∑_r=1^r=n  1/n(r/n + √r/n)

now replace 1/n by dx and r/n by x

limit of integral will be r/n

lower limit n tend to ∞ and r=1 will be 0

upper limit for r=n will be 1

it will convert into integral ∫₀¹ dx/(x+√x) = ∫₀¹dx/√x(1+√x)

now (1+√x)=t

and dx/2√x = dt

lower limit for x=0 , t=1 andupper limit for x=1, t=2

integral become ∫₁² 2dt/t= 2(lnt)₁²= 2ln2=ln4 answer

All the best.                                                           

AKASH GOYAL

AskiitiansExpert-IITD

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