Let `S_n=sum_(k=0)^n n/(n^2+k n+k^2) and T_n=sum_(k=0)^(n-1)n/(n^2+k n+k^2)`,for `n=1,2,3,.......,` thenA. `S_(n)lt(pi)/(3sqrt(3))`B. `S_(n)gt(pi)/(3sqrt(3))`C. `T_(n) lt (pi)/(3sqrt(3))`D. `T_(n)gt(pi)/(3sqrt(3))`

Let `S_n=sum_(k=0)^n n/(n^2+k n+k^2) and T_n=sum_(k=0)^(n-1)n/(n^2+k n

1.	Let `S_n=sum_(k=0)^n n/(n^2+k n+k^2) and T_n=sum_(k=0)^(n-1)n/(n^2+k n+k^2)`,for `n=1,2,3,.......,` thenA. `S_(n)lt(pi)/(3sqrt(3))`B. `S_(n)gt(pi)/(3sqrt(3))`C. `T_(n) lt (pi)/(3sqrt(3))`D. `T_(n)gt(pi)/(3sqrt(3))`
Answer» Correct Answer - A::D `S_(n)lt lim_(nto oo) S_(n)=lim_(n to oo) sum_(k=1)^(n)1/n 1/(1+k//n+(k//n)^(2))` `=int_(0)^(1)(dx)/(1+x+x^(2))` `=int_(0)^(1)(dx)/((x+1/2)^(2)+3/4)` `=[2/(sqrt(3))"tan"^(-1)((x+1/2)/((sqrt(3))/2))]_(0)^(1)=(pi)/(3sqrt(3))` Now `T_(n)gt(pi)/(3sqrt(3))`as `h sum_(k=0)^(n-1)f(k//n)gt int_(0)^(1)f(x)dxgt h sum_(k=1)^(n)f(k//n)`

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
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`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`