Inequalities in integration We are given the value of $S_n$ below and we have to find if $S_n$ is greater or smaller than $\dfrac{\pi}{3\sqrt{3}}$.
$$S_n=\sum\limits_{k=1}^n \frac{n}{n^2+kn+k^2}$$
$$S_n<\dfrac{\pi}{3\sqrt{3}} \quad \text{or}\quad S_n>\dfrac{\pi}{3\sqrt{3}}$$
I tried it as follows:
$$S_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$$
$$\rightarrow \frac{1}{x^2+x+1}$$
But at last I thought to draw the graph so that I can do something with area under the curve or can thought at extreme points i.e at $n$ is approaching to infinity.
I got stuck not able to do anything. You can see the image of my attempt here.


Look at this picture. The red curve is the function $\displaystyle f(x)=\frac{1}{x^2+x+1}$
The red curve is decreasing on $[0,1]$, since $\displaystyle f'(x)=-\frac{2x+1}{\left(x^2+x+1\right)^2}<0$ for $0\leq x \leq 1$.
The gray shaded area is the value of $\displaystyle\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$ when $n=6$. 
Can you see why? The width of each box is $\displaystyle \frac{1}{n}$, and the height of the $k$th box is $\displaystyle f\left(\frac{k}{n}\right)=\frac{1}{\left(\frac{k}{n}\right)^2+\frac{k}{n}+1}$.
As you can see, as $n$ increases, the number of boxes will increase, but the shaded area will never be equal to or greater than the area under the red curve for the interval $[0,1]$. (By the way, these approximations of the area under the curve are called Riemann sums.)
Ergo, we conclude that $\displaystyle S_n<\int_{0}^{1}\frac{dx}{x^2+x+1}$
$$\begin{align}
\int_{0}^{1}\frac{dx}{x^2+x+1}&=\int_{0}^{1}\frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}
\\
&=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{du}{u^2+1} \qquad\text{(Substitute $u=\frac{2x+1}{\sqrt{3}}$, $\frac{du}{dx}=\frac{2}{\sqrt{3}}$)}\\
&=\frac{2}{\sqrt{3}}\arctan(u)\bigg|_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\\
&=\frac{\pi}{3\sqrt{3}}
\end{align}$$
And thus we have our desired result. 
(By the way, since I am bad at calculus, I used the following sites: http://www.integral-calculator.com/, http://www.derivative-calculator.net/)
