2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Taking n=1 last inequality proves to be correct $\endgroup$ – Loop Back Dec 12 '18 at 8:24
  • $\begingroup$ Are you talking about my "last inequality" or yours? $\endgroup$ – Robert Z Dec 12 '18 at 8:34
  • $\begingroup$ Yes............... $\endgroup$ – Loop Back Dec 12 '18 at 8:39
  • $\begingroup$ @LoopBack Yes what? $\endgroup$ – Robert Z Dec 12 '18 at 8:42
  • $\begingroup$ My last inequality is correct for n=1 $\endgroup$ – Loop Back Dec 12 '18 at 8:43
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.