Let `S_(n)=sum_(k=1)^(n)n/(n^(2)+kn+k^(2))` and `T_(n)=sum_(k=0)^(n-1)n/(n^(2)+kn+k^(2))` for `n=1,2,3,`………. Then

Options

A. `S_(n) lt(pi)/(3sqrt(3)),T_(n)gt(pi)/(3sqrt(3))`
B. `S_(n) ge(pi)/(3sqrt(3)),T_(n)lt(pi)/(3sqrt(3))`
C. `S_(n) gt(pi)/(3sqrt(3)),T_(n)gt(pi)/(3sqrt(3))`
D. `S_(n) lt(pi)/(3sqrt(3)),T_(n)lt(pi)/(3sqrt(3))`

Answer

A

Explanation

`S_(n)=underset(k=1)overset(n)sum (n)/(n^(2)+nk+k^(2))`
rArr`S_(n)(1)/(2)=underset(k=1)overset(n)sum (1)/(1+(k)/(n)+(k^(2))/(n^(2)))`
rArr`S_(n)ltlim_(n to oo)underset(k=1)overset(n)sum{ (1)/(1+(k)/(n)+((k^(2))/(n^(2)))^(2))}(1)/(n)`
`rArr S_(n)lt underset(0)overset(1) int (1)/(1+x+x+x^(2))dx`
`rArr S_(n)lt underset(0)overset(1) int (1)/((x(1)/(2))^(2)+(sqrt(3)/(2))^(2))dx`
`rArr S_(n)lt(2)/(sqrt(3))[ "tan" ^(-1)(2x+1)/(sqrt(3))]_(0)^(1)`
`rArr S_(n)(2)/(sqrt(3))xx((pi)/(3)-(pi)/(6))rArr S_(n)lt (pi)/(3sqrt(3))`
As `f(x)=(1)/(1+x+x^(2))` is decreasing function on [0,1].
`:. h underset(k=0) overset(n-1)sumf(kh)gtunderset(0)overset(1)intf(x)dxgthunderset(k=1) overset(n)sum(kh)`
`rArr T_(n) gt(pi)/(3sqrt(3))gtS_(n)`