2
$\begingroup$

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

Now the limit is very easy to go using the sandwich theorem which yields in the answer as $\frac 13$. But I want to know if there is any corresponding solution which inturn uses Riemann Sums and Integrals to evaluate this limit.

Edit:

Another such question is $$\lim_{n\to \infty} \sum_{r=1}^n \frac {n}{n^2+r}$$

It could have been much easier to solve if the above question had been $$\lim_{n\to \infty} \sum_{r=1}^n \frac {n}{n^2+r^2}$$ which simplifies to $$\int_0^1 \frac {dx}{1+x^2}=\frac {\pi}{4}$$

$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

$$\sum_{r=1}^{n}\frac{r^2}{n^3+r^2}\leq \frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2 \to \int_{0}^{1}x^2\,dx = \frac{1}{3} $$ and $$\sum_{r=1}^{n}\frac{r^2}{n^3+r^2}\geq \frac{n^3}{n^3+n^2}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2 \to \frac{1}{3}.$$ Essentially, the $r^2$ term in the denominator is negligible (with or without Riemann sums).
We cannot state the same for $$ \lim_{n\to +\infty}\sum_{r=1}^{n}\frac{r^2}{n^3+r^{\color{red}{3}}}=\lim_{n\to +\infty}\frac{1}{n}\sum_{r=1}^{n}\frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3}=\int_{0}^{1}\frac{x^2\,dx}{1+x^3}=\frac{\log 2}{3}. $$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.