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

510 Points
```										 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...

```
6 years ago
PRATIK subramaniyam SAXENA
30 Points
```										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.
```
6 years ago
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