Let we are given summation as ${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}$ and ${{T}_{n}}=\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}$ for $n=1,2,3...$ then;\[\]
A.${{S}_{n}}<\dfrac{\pi }{3\sqrt{3}}$ \[\]
B. ${{S}_{n}}>\dfrac{\pi }{3\sqrt{3}}$\[\]
C.${{T}_{n}}<\dfrac{\pi }{3\sqrt{3}}$\[\]
D.${{T}_{n}}>\dfrac{\pi }{3\sqrt{3}}$\[\]

Hint: We are going to use Riemann integral as a limit of sum. We take limit $n\to \infty $ on the summation ${{S}_{n}}$ and deduce that ${{S}_{n}}<\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$. We express ${{S}_{n}}$ in the form of $\sum\limits_{k=1}^{n}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}$. We put $\dfrac{k}{n}=x$ and the use the Riemann integral formula $\int_{0}^{1}{f\left( x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}$. We similarly proceed for ${{T}_{n}}$ where we use the Riemann integral formula $\int_{0}^{1}{f\left( x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=0}^{n-1}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}$.\[\]

Complete step-by-step solution
We know from Riemann’s integration that we can convert the limit of a sum to definite integral as
\[\int_{0}^{1}{f\left( x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{n-1}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}\]
We are given two summations in the question as;
\[\begin{align}
  & {{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}} \\
 & {{T}_{n}}=\sum\limits_{k=1}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}} \\
\end{align}\]
We see in the options the bounds of ${{S}_{n}}$ and ${{T}_{n}}$. So let us consider the limit $n\to \infty $ on ${{S}_{n}}$ an and have
\[{{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}<\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}\]
Let take ${{n}^{2}}$ common from the denominator terms in the right hand side to have;
\[\begin{align}
  & \Rightarrow {{S}_{n}}<\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}\left( 1+\dfrac{k}{n}+\dfrac{{{k}^{2}}}{{{n}^{2}}} \right)}} \\
& \Rightarrow {{S}_{n}}<\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{1}{n}\dfrac{1}{\left( 1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}} \right)}} \\
\end{align}\]
We assign variable $\dfrac{k}{n}=x$ and use Riemann’s integration formula to express the summation as definite integral as
\[\Rightarrow {{S}_{n}}<\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}}.......\left( 1 \right)\]
We express the polynomial in the denominator in terms of complete square so that we can use the standard indefinite integral $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C},a\ne 0$. So we have;
\[\begin{align}
  & \Rightarrow 1+x+{{x}^{2}} \\
 & \Rightarrow {{x}^{2}}+2\cdot x\cdot \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}+1 \\
 & \Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4} \\
 & \Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\
\end{align}\]
We put the above obtained expression in (1) to have;
\[\Rightarrow {{S}_{n}}<\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}\]
We use the standard integral $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C},a\ne 0$ to evaluate the definite integral within limits have;
\[\begin{align}
  & \Rightarrow {{S}_{n}}<\left[ \dfrac{1}{\dfrac{\sqrt{3}}{2}}{{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \right) \right]_{x=0}^{x=1} \\
 & \Rightarrow {{S}_{n}}<\dfrac{2}{\sqrt{3}}\left[ {{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right) \right]_{x=0}^{x=1} \\
 & \Rightarrow {{S}_{n}}<\dfrac{2}{\sqrt{3}}\left[ {{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\dfrac{1}{\sqrt{3}}\right] \\
 & \Rightarrow {{S}_{n}}<\dfrac{2}{\sqrt{3}}\left( \dfrac{\pi }{3}-\dfrac{\pi }{6} \right)=\dfrac{\pi }{3\sqrt{3}}....\left( 1 \right) \\
\end{align}\]
So we have the upper bounds for ${{S}_{n}}$ as ${{S}_{n}}<\dfrac{\pi }{3\sqrt{3}}$. We can similarly take limit $n\to \infty $ for the summation of ${{T}_{n}}$ and then take ${{n}^{2}}$ common in the denominator of ${{T}_{n}}$ to have;
\[\begin{align}
  & {{T}_{n}}=\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}<\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}} \\
 & \Rightarrow {{T}_{n}}<\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=0}^{n-1}{\dfrac{1}{n}\cdot \dfrac{1}{1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}}}} \\
\end{align}\]
We similarly take $x=\dfrac{k}{n}$ in the above step and then use Riemann’s integration formula to have;
\[\Rightarrow {{T}_{n}}<\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}}\]
We have already evaluated the above definite integral as $\dfrac{\pi }{3\sqrt{3}}$. So we have;
\[\Rightarrow {{T}_{n}}<\dfrac{\pi }{3\sqrt{3}}.......\left( 2 \right)\]
So we see from inequalities (1) and (2) that the correct options are A and C.

Note: We note that $n$ is the number of divisions that divides the interval $\left[ 0,1 \right]$ which is also the bounds of the integral. We can find these limits by taking $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{r}{n}$ with the initial and final value of $r$in first and last term of the summation. The general equation for definite integral as a limit of sum is given by $\int_{a}^{b}{f\left( x \right)}=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{n} \right)\left[ f\left( a \right)+f\left( a+h \right)...+f\left\{ a+\left( n-1 \right) \right\}h \right]$ where $h=\dfrac{b-a}{n}$.