# 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.

• Should the third formula be $\displaystyle \lim_{ n \to \infty} \sum_{r=1}^n\frac{r(\dfrac rn+\dfrac 1n)}{\dfrac rn-n^2}$ ? Commented Jul 8, 2021 at 8:30
• You can ignore any terms that come out to things like $\frac{k}{n^2}$, $\frac{k}{n^3}$, or higher because their contribution will vanish more rapidly than the leading order contributions. This will simplify to $\sum -\frac{k^2}{n^2}\cdot\frac{1}{n} \to \int_0^1 -x^2dx$. You can prove this with squeeze theorem. Commented Jul 8, 2021 at 8:32
• @TitoEliatron Thankyou, I assure you that was just a typo. Commented Jul 8, 2021 at 8:34
• @NinadMunshi If I further divide numerator and denominator by $n^2$, I get $$\lim_{ n \to \infty} \sum_{r=1}^n\frac{\dfrac rn(\dfrac r{n^2}+\dfrac1{n^2})}{\dfrac r{n^3}-1}$$ So, here, $\dfrac r{n^3}$ and $\dfrac1{n^2}$ would be ignored? So, yes, that's indeed giving me $$\int_0^1-x^2dx$$ So, the answer is $-\dfrac13$. Thankyou. Commented Jul 8, 2021 at 8:48
• @NinadMunshi While two beautiful answers have already been posted, any chance you could post your comment as answer? I'll accept it. Thanks. Commented Jul 8, 2021 at 8:56

$$R=\lim_{ n \to \infty} \left(\frac{1^2+1}{-n^3}+\frac{2^2+2}{-n^3}+\frac{3^2+3}{-n^3}+...+\frac{n^2+n}{-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{-6n^3}\right)=\frac{-1}{3}$$
$$P=\lim_{ n \to \infty} \left(\frac{1^2+1}{n-n^3}+\frac{2^2+2}{n-n^3}+\frac{3^2+3}{n-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{6(n-n^3)}\right)=\frac{-1}{3}$$
$$R\ge\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)\ge P$$
So $$~\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)=\frac{-1}{3}$$
Hint: $$k-n^{3}$$ lies between $$1-n^{3}$$ and $$n-n^{3}$$ for $$1 \leq k \leq n$$. This allows you to show that $$\lim \sup$$ and $$\lim \inf$$ are both equal to $$-\frac 1 3$$ using the formulas for $$\sum\limits_{k=1}^{n} k$$ and $$\sum\limits_{k=1}^{n} k^{2}$$.