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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.
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...
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.
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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