Limit of sum as definite integral I don't understand why $$\displaystyle \sum_{k=1}^n \dfrac{n}{n^2+kn+k^2} < \lim_{n\to \infty}\sum_{k=1}^n \dfrac{n}{n^2+kn+k^2}$$
whereas
$$\displaystyle \sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} > \lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} $$
I know that $$\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\dfrac{\pi}{3\sqrt{3}}$$ $$\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}=\dfrac{\pi}{3\sqrt{3}}$$
I saw somewhere on the internet that $$\displaystyle \dfrac{1}{n}\sum_{k=0}^{n-1} f\left(\dfrac{k}{n}\right) > \int_0^1 f(x)dx >  \dfrac{1}{n}\sum_{k=1}^{n} f\left(\dfrac{k}{n}\right)$$ Why is this true?
 A: Consider the function
$$f(x)=\frac{1}{1+x+x^2}$$
and note that the your inequality holds because $f$ is strictly decreasing. 
Indeed for $n\geq 1$, $k\geq 0$, and  $x\in [\frac{k}{n},\frac{k+1}{n}],$
$$f(\frac{k}{n})> f(x)> f(\frac{k+1}{n}).$$
By integrating over the interval $[\frac{k}{n},\frac{k+1}{n}]$, we get
$$\frac{f(\frac{k}{n})}{n}=\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(\frac{k}{n})dx> \int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx> \int_{\frac{k}{n}}^{\frac{k+1}{n}}f(\frac{k+1}{n})dx=\frac{f(\frac{k+1}{n}) }{n}.$$
Finally we take the sum for $k=0,\dots,n-1$,
$$\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})>\int_0^1 f(x)\,dx
>\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k+1}{n}).$$
Note that the last sum on the right is equal to 
$$\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})=
\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})+\frac{f(1)-f(0)}{n}.$$
P.S. Once we have the double inequality, we may conclude that
$$\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}=\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\int_0^1\frac{dx}{1+x+x^2}=\dfrac{\pi}{3\sqrt{3}}.$$
A: In this case $f(x)=\frac{1}{1+x+x^2}$ is decreasing on $[0,\,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.
