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

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.

  • 7
    $\begingroup$ 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. $\endgroup$ – r9m Dec 28 '14 at 10:37
  • $\begingroup$ 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 $\approx \frac{3}{4}+\frac{1}{2 n}$ $\endgroup$ – Claude Leibovici Dec 28 '14 at 10:38
  • $\begingroup$ See also: math.stackexchange.com/questions/478344/…, math.stackexchange.com/questions/150391/… and other similar questions. $\endgroup$ – Martin Sleziak Dec 29 '14 at 10:33

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

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

  • $\begingroup$ I understand the steps here. I guess this becomes a Riemann sum, then. $\endgroup$ – Berk Özbalcı Dec 28 '14 at 10:45

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.


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

  • $\begingroup$ Can we convert the sum to an integral like this? Why does this work? $\endgroup$ – Berk Özbalcı Dec 28 '14 at 10:35
  • $\begingroup$ A high school stıdent is familair with riemann sum ? $\endgroup$ – mesel Dec 28 '14 at 10:36
  • $\begingroup$ @TheConjuring: Are you familiar with Riemann sum ? $\endgroup$ – mesel Dec 28 '14 at 10:38
  • $\begingroup$ there is anyway no other possibility than using Riemann sum ! $\endgroup$ – idm Dec 28 '14 at 10:40
  • $\begingroup$ I know what a Riemann sum is, but I couldn't recognize it. Neat, though. $\endgroup$ – Berk Özbalcı Dec 28 '14 at 10:42

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