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$$ \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} $$

High school student here! This was a question from our Mathematics exam (prior to learning derivatives). Now there was some sort of a bounty here in our school, but nobody could solve it even after weeks passed.

WolframAlpha gives $3/4$ but no other explanation. I'm curious how one could tackle the expression in the numerator.

DeepSea
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Berk Özbalcı
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    Hint: $$\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} = \lim_{n \to \infty} \frac{(n+1)^{1/3}}{(n+1)^{4/3} - n^{4/3}}$$ by Stolz-Cesaro Theorem. – r9m Dec 28 '14 at 10:37
  • You will learn later that the numerator is the harmonic number $H_n^{\left(-\frac{1}{3}\right)}$ and, for large values of $n$, the expression is $\approx \frac{3}{4}+\frac{1}{2 n}$ – Claude Leibovici Dec 28 '14 at 10:38
  • See also: http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk1, http://math.stackexchange.com/questions/150391/evaluate-lim-limits-n-to-infty-frac-sum-k-1n-kmnm1 and other similar questions. – Martin Sleziak Dec 29 '14 at 10:33

3 Answers3

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$$\frac{1^{1/3}+2^{1/3}+...+n^{1/3}}{n\cdot n^{1/3}}=\frac{\sum_{k=1}^n k^{1/3}}{n\cdot n^{1/3}}=\frac{n^{1/3}}{n \cdot n^{1/3}}\sum_{k=1}^n\left(\frac{k}{n}\right)^{1/3}=\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^{1/3}$$ Then,

$$\lim_{n\to\infty }\frac{1^{1/3}+2^{1/3}+...+n^{1/3}}{n\cdot n^{1/3}}=\lim_{n\to\infty }\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^{1/3}=\int_0^1 x^{1/3}dx=...$$

idm
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Consider that, since $a-b=\frac{a^3-b^3}{a^2+ab+b^2},$ $$ (n+1)^{\frac{4}{3}}-n^{\frac{4}{3}} = \frac{(n+1)^4-n^4}{3n^{\frac{8}{3}}+O\left(n^{\frac{5}{3}}\right)}=\frac{4}{3}n^{\frac{1}{3}}+O\left(n^{-\frac{2}{3}}\right).\tag{1}$$ By multiplying both sides of the previous equation by $\frac{3}{4}$, then summing over $n$ we get that the limit is $\color{red}{\frac{3}{4}}$, as expected. This is just an application of creative telescoping and trivial inequalities.

Jack D'Aurizio
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Hint: $\displaystyle \int_{0}^1 x^{\frac{1}{3}}dx= 3/4$. The numerator is a Riemann sum !

DeepSea
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