Calculate ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $ Let ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $, for n=1,2,3,.....Then
(A) ${S_n} < \frac{\pi }{{3\sqrt 3 }}$
(B) ${S_n} > \frac{\pi }{{3\sqrt 3 }}$
(C) ${T_n} < \frac{\pi }{{3\sqrt 3 }}$
(D) ${T_n} > \frac{\pi }{{3\sqrt 3 }}$
The official Answer is A and D
My approach is as follows
$\mathop {\lim }\limits_{n \to \infty } {S_n} = \sum\limits_{k = 1}^n {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
${S_n} < \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + 1}}}  = \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + \frac{1}{4} + \frac{3}{4}}}}  = \int\limits_0^1 {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{1}{{\frac{{\sqrt 3 }}{2}}}} \left( {\left. {{{\tan }^{ - 1}}\frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right|_0^1} \right) = \frac{2}{{\sqrt 3 }}\left( {{{\tan }^{ - 1}}\sqrt 3  - {{\tan }^{ - 1}}\frac{1}{{\sqrt 3 }}} \right) = \frac{2}{{\sqrt 3 }}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{3\sqrt 3 }}$
So A is correct
$\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
$t = k + 1\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{t = 1}^n {\frac{1}{{n\left( {1 + \frac{{t - 1}}{n} + {{\left( {\frac{{t - 1}}{n}} \right)}^2}} \right)}}} $
The answer is D but don't know to proceed.
 A: This is a beautiful question asked in JEE Advanced which tests your basic knowledge about definite integrals- Riemann Sum.

We have:
$${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $$
Note that both these functions are decreasing.
Therefore
$\lim_{n\to\infty}\sum_{k = 1}^n f(x)$ represents left Riemann sum
and
$\lim_{n\to\infty}\sum_{k = 0}^{n-1} f(x)$ the right Riemann sum.

 
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Left Riemann Sum $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Right Riemann Sum

Now show that the value of integral is $\boxed{\frac{\pi}{3\sqrt3}}$
Conclude that $T_n$ is greater than this and $S_n$ is lesser than this
A: We have $$S_n \le I \le T_n$$ because $f(x)$ here is a decreasing function.
By squeez law, we have
$$I=\lim_{n \to \infty} T_n=\lim_{n \to \infty} S_n=\int_{0}^{1} f(x) dx $$
$$f(x)=\frac{1}{x^2+x+1}$$
So $$I=\int_{0}^{1} \frac{dx}{(x+1/2)^2+3/4}=\frac{2}{\sqrt{3}}\tan^{=1}\frac{(2x+1)}{\sqrt{3}}=\frac{2}{\sqrt{3}}[\pi/3-\pi/6]=\frac{\pi}{3\sqrt{3}}$$
So options (A) and (B) are correct.
