Let $S_n=\sum\limits_{k=1}^n \frac{n}{n^2+n k+k^2}$ and $T_n=\sum\limits_{k=0}^{n-1} \frac{n}{n^2+n k+k^2}$, for $n=1,2,3, ....$, then

$S_n<\frac{\pi}{3 \sqrt{3}}, T_n>\frac{\pi}{3 \sqrt{3}}$

We have,

$S_n =\sum\limits_{k=1}^n \frac{n}{n^2+n k+k^2}$

$\Rightarrow S_n =\frac{1}{n} \sum\limits_{k=1}^n \frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}$

$\Rightarrow S_n<\lim\limits_{n \rightarrow \infty} \sum\limits_{k=1}^n\left\{\frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^2}\right\} \frac{1}{n}$

$\Rightarrow S_n<\int\limits_0^1 \frac{1}{1+x+x^2} d x$

$\Rightarrow S_n<\int\limits_0^1 \frac{1}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d x$

$\Rightarrow S_n<\frac{2}{\sqrt{3}}\left[\tan ^{-1} \frac{2 x+1}{\sqrt{3}}\right]_0^1$

$\Rightarrow S_n<\frac{2}{\sqrt{3}} \times\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \Rightarrow S_n<\frac{\pi}{3 \sqrt{3}}$

As $f(x)=\frac{1}{1+x+x^2}$ is a decreasing function on [0, 1]

∴  $h \sum\limits_{k=0}^{n-1} f(k h)>\int\limits_0^1 f(x) d x>h \sum\limits_{k=1}^n f(k h)$

$\Rightarrow T_n>\frac{\pi}{3 \sqrt{3}}>S_n$