# 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?

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 $$\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}=\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\int_0^1\frac{dx}{1+x+x^2}=\dfrac{\pi}{3\sqrt{3}}.$$

• Taking n=1 last inequality proves to be correct – Loop Back Dec 12 '18 at 8:24
• Are you talking about my "last inequality" or yours? – Robert Z Dec 12 '18 at 8:34
• Yes............... – Loop Back Dec 12 '18 at 8:39
• @LoopBack Yes what? – Robert Z Dec 12 '18 at 8:42
• My last inequality is correct for n=1 – Loop Back Dec 12 '18 at 8:43

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.