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

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vikas askiitian expert · Answer

summation r/[4r 2 + 1] limit n-> infinity summation (r) n(n+1)/2 ( first n natural numbers) summation (r 2 ) 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) + 3y 2 ] 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...

PRATIK subramaniyam SAXENA · Answer

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.

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

Evaluate-

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

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