Limit of sum as definite integral

Question

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?

Your last inequality should be reversed!! See my answer below. — Robert Z, Dec 12 '18 at 08:34
Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums. — user, Dec 12 '18 at 09:11

Robert Z · Accepted Answer · 2018-12-12T09:13:27.917

1

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

edited Dec 12 '18 at 09:13

answered Dec 12 '18 at 07:56

Robert Z

147,345

Taking n=1 last inequality proves to be correct – Dec 12 '18 at 08:24
Are you talking about my "last inequality" or yours? – Robert Z Dec 12 '18 at 08:34
Yes............... – Dec 12 '18 at 08:39
@LoopBack Yes what? – Robert Z Dec 12 '18 at 08:42
My last inequality is correct for n=1 – Dec 12 '18 at 08:43
Sorry, my fault. You are right! – Robert Z Dec 12 '18 at 08:45
I m unable to understand why you have taken x€(k/n , k+1/n) – Dec 12 '18 at 08:46
And how is $$\frac{f(\frac{k}{n})}{n}=\int_{\frac{k}{n}}^{\frac{k}{n}}f(\frac{k+1}{n})dx$$ – Dec 12 '18 at 08:47
In order to show the inequality I first consider what happens in each subinterval $[k/n,(k+1)/n]$ of $[0,1]$. – Robert Z Dec 12 '18 at 08:47
That's a typo, sorry again. – Robert Z Dec 12 '18 at 08:49
btw is this expression correct $$\lim_{n\to \infty}\sum_{k=1}^{n} \dfrac{n}{n^2+kn+k^2}=\lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2}$$ – Dec 12 '18 at 09:06
Yes, it is correct. – Robert Z Dec 12 '18 at 09:08
Finally I understand what you are trying to prove. Actually this result is general. For decreasing functions this is true $$\displaystyle \dfrac{1}{n}\sum_{k=0}^{n-1} f(\dfrac{k}{n}) > \int_0^1 f(x)dx > \dfrac{1}{n}\sum_{k=1}^{n} f(\dfrac{k}{n})$$ whereas this is true for increasing functions $$\displaystyle \dfrac{1}{n}\sum_{k=0}^{n-1} f(\dfrac{k}{n}) < \int_0^1 f(x)dx < \dfrac{1}{n}\sum_{k=1}^{n} f(\dfrac{k}{n})$$ isn't it? – Dec 12 '18 at 09:13
Yes. they are correct. – Robert Z Dec 12 '18 at 09:14
Boo yeah. Thanks a lot. After reading this answer 1000 times. I finally get it – Dec 12 '18 at 09:15

score 1 · Answer 2 · answered Dec 12 '18 at 07:57

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.

Limit of sum as definite integral

2 Answers2