# Limit of $\lim\limits_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$

I want to evaluate this limit. $$\lim_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$$ What I did is:
set $f(x)=x^\frac{1}{3}$ $$f(1)+\int^\infty_0 x^\frac{1}{3} \, dx<f(1)+f(2)+\dots+f(n)<f(n)+\int^\infty_0 x^\frac{1}{3}\,dx$$ how it can help me to evaluate this limit?
I can convert the right\left expressions to limits? or the whole inequality to expression of limit.
I need some advice here, Thanks!

• It should be compared with $\int_0^1 x^{\frac 1 3}dx$ – Sungjin Kim Jun 8 '13 at 16:14
• Did you mean $n\to\infty$ rather than $x\to\infty$? – Michael Hardy Jun 8 '13 at 16:16
• The right idea. You don't want to integrate to $\infty$. More like $n$. – André Nicolas Jun 8 '13 at 16:16
• its ok to do : $$\lim_{n\to \infty} \frac{n^{1/3}+\frac{3}{4}n^{4/3}}{n^{4/3}} = \frac{\infty}{\infty} \rightarrow L`hopital =$$ after calculations $\frac{3}{4}$? – Ofir Attia Jun 8 '13 at 16:29
• Try to interpret it as a Riemann sum. – jdoicj Jun 8 '13 at 16:39

Hint: An integral over $[0,\infty)$ does not look right to me, it seems better to observe that with your $f$ as above \begin{align*} \frac{1^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} &= \frac{f(1/n) + f(2/n) + \cdots + f(1)}{n}\\ &= \sum_{i=1}^n f(i/n) \cdot \frac 1n \end{align*}
It is easy to show that $$\sum_{k=1}^n k^a = \begin{cases}\dfrac{n^{a+1}}{a+1} + \mathcal{O}(n^a) & \text{if } a \neq -1\\ \log(n) + \mathcal{O}(1) & \text{if }a = - 1\end{cases}$$ Conclude from this.
We have $$\sum_{k=1}^n k^{\frac{1}{3}}\sim_\infty\int_1^nx^{\frac{1}{3}}dx\sim_\infty\frac{3}{4}n^{\frac{4}{3}}$$ so $$\lim_{x\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)=\frac{3}{4}$$