Evaluate-

summation r = 1 to r =n , r/[4(r^2)+1],when n tends to infinite.

summation r/[4r² + 1] limit n-> infinity

summation (r) n(n+1)/2                  ( first n natural numbers)

summation (r²) n(n+1)(2n+1)/6     (sum of squares of first n nutural numbers)

so , summation r = 1 to r =n , r/[4(r^2)+1]

               = 3(n(n+1))/2[2(n(n+1)(2n+1) + 3]   lim n-> infinity

 put , n = 1/y

                =3(y+1)/2[2(y+1)(y+2) + 3y²]   limit y->0

now on taking limit

               = 3(1)/2[4] = 3/8

so , the value is 3/8 ...

approve if u like my ans...

SORRY !!!!!!!! The answer is 1/4

secondly u have done wrong, as i know we cant put the direct values of sum of n terms in the denominator .

we make a series and then try to solve it. but i did like that but it became very tough.