<s>[INST] Find the value of $\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$ The following question is taken from the practice set of JEE Main exam.

Find the value of $$\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$$

I wrote it as$$\lim_{ n \to \infty} \sum_{r=1}^n\frac{r^2+r}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(r+1)}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(\dfrac rn+\dfrac1n)}{\dfrac rn-n^2}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{\dfrac rn(\dfrac rn+\dfrac1n)n}{\dfrac rn-n^2}$$
To convert it into into integration, $\dfrac rn$ can be written as $x$. $\dfrac1n$ is written as $dx$, but instead we have $n$. How to tackle that? Or any other approach for the question? Just a hint would suffice. Thanks.
 [/INST] $R=\lim_{ n \to \infty} \left(\frac{1^2+1}{-n^3}+\frac{2^2+2}{-n^3}+\frac{3^2+3}{-n^3}+...+\frac{n^2+n}{-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{-6n^3}\right)=\frac{-1}{3}$
$P=\lim_{ n \to \infty} \left(\frac{1^2+1}{n-n^3}+\frac{2^2+2}{n-n^3}+\frac{3^2+3}{n-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{6(n-n^3)}\right)=\frac{-1}{3}$
$$R\ge\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)\ge P$$
So $~\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\frac{-1}{3}$
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