To evaluate the limits $\lim\limits_{n \to \infty} \{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}$

Question

To me it seems like that we need to manipulate the given sum into the Riemann sum of some function. First writing in the standard summation form;

$$\{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}=\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}$$

I have tried factoring out $\frac{1}{n}$ from the above sum. But I cant seem to get into the right form. Any help!

Answer 1

You can proceed from here as $$\lim_{n\to \infty}\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}= \lim_{n\to\infty}\sum\limits_{k=0}^n \;\frac{1}{n}\frac{\frac{k^2}{n^2}}{\frac{k^3}{n^3}+1}= \lim_{h\to 0} \;h\sum_{k=1}^n \frac{{(kh)}^2}{1 + {(kh)}^3}= \int_0^1 \frac{x^2}{1+x^3} \mathrm{d}x\;.$$

Answer 2

Note that $$\frac{k^2}{k^3+n^3}=\dfrac1n\cdot\dfrac{\left( \frac{k}{n} \right)^2}{\left( \frac{k}{n} \right)^3+1}$$ so you can rewrite the summation to be $$\begin{align}\lim_{n\to \infty} \Big(\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\Big) &= \lim_{n\to \infty} \sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}\\ &= \lim_{n\to \infty} \sum\limits_{k=0}^n \dfrac1n\cdot\dfrac{\left( \frac{k}{n} \right)^2}{\left( \frac{k}{n} \right)^3+1}\\ &= \lim_{h\to 0} \;h\sum_{k=1}^n \frac{{(kh)}^2}{1 + {(kh)}^3}\\ &= \int_0^1 \frac{x^2}{1+x^3} \mathrm{d}x\;\\ &= \frac{1}{3} \ln(1+x^3) \Big|_0^1\\ &= \frac{1}{3} \ln 2 \end{align}$$