$\lim_{n \to \infty}\sum_{r=1}^{n}\frac{r^2}{n^3 + n^2 + r}$

Question

$$\lim_{n \to \infty}\displaystyle\sum_{r=1}^{n}\dfrac{r^2}{n^3 + n^2 + r}= ?$$

I tried using Newton Leibnitz Formula to map $\frac rn$ to $x$ and $\frac 1n$ to $dx$ but that doesn't help here. How to solve the problem then?

Answer 1

Note that $$\dfrac{1}{1 + \frac{1}{n}+\frac{1}{n^2}}\cdot S_n=\sum_{r=1}^{n}\dfrac{r^2}{n^3 + n^2+n}\leq \sum_{r=1}^{n}\dfrac{r^2}{n^3 + n^2+r}\leq \sum_{r=1}^{n}\dfrac{r^2}{n^3}=S_n$$ where $S_n=\frac{1}{n}\sum_{r=1}^{n}(r/n)^2$ is a Riemann sum.

Are you able to find $\lim_{n\to \infty} S_n$? What is the desired limit?

Answer 2

Let $$ a_n=\sum_{r=1}^{n}\dfrac{r^2}{n^3 + n^2 + r}=\frac{1}{n}\sum_{r=1}^{n}\dfrac{\left(\frac{r}{n}\right)^2}{1 + \frac1n + \frac{r}{n^2}}. $$ Then observe that $$ \frac{1}{1+\frac2n}\cdot\frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2=\frac{1}{n}\sum_{r=1}^{n}\dfrac{\left(\frac{r}{n}\right)^2}{1 + \frac1n + \frac{n}{n^2}}\le a_n\le \frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2 $$ But (Riemann sum) $$ \frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2\to\int_0^1 x^2\,dx=\frac{1}{3}, $$ and hence $$ a_n\to \frac13. $$

Answer 3

We have that

$$\dfrac{r^2}{n^3 + n^2 + r}=\frac1{n^3}\dfrac{r^2}{1 + 1/n + r/n^3}=\frac{r^2}{n^3}(1 + 1/n + r/n^3)^{-1}=\frac{r^2}{n^3}+r^2O\left(\frac{1}{n^4}\right)$$

then recall that

$$\sum_{r=1}^{n} r^2=\frac{(n)(n+1)(2n+1)}{6}$$

and thus

$$\sum_{r=1}^{n}\dfrac{r^2}{n^3 + n^2 + r}=\left(\frac{1}{n^3}+ O\left(\frac{1}{n^4}\right)\right)\sum_{r=1}^{n}r^2\\=\frac{(n)(n+1)(2n+1)}{6n^3}+\frac{(n)(n+1)(2n+1)}{6}O\left(\frac{1}{n^4}\right)\to\frac13$$