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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!

Martin Sleziak
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Miz
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    You can use sandwich law. What is the biggest numerator? and what is the smallest denominator? How can you bound the sum? – Galc127 Jan 19 '16 at 09:16
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    See http://math.stackexchange.com/questions/1279816/find-the-limit-lim-n-rightarrow-infty-sum-k-1n-frac1kn and observe : $$\frac{k^2}{k^3+n^3}=\dfrac1n\cdot\dfrac{(k/n)^2}{(k/n)^3+1}$$ – lab bhattacharjee Jan 19 '16 at 09:17
  • @labbhattacharjee That seems to do it. The answer turns out to be $\int_{0}^1 \frac{x^2}{1+x^3} = \frac{1}{3} \log 2$. – Miz Jan 19 '16 at 09:27
  • @lab bhattacharjee: Hmmm... It seems I'm very slow in using mathjax :( Please turn your comment as an answer; I'll delete mine. –  Jan 19 '16 at 09:33

2 Answers2

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

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

sirfoga
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