How can I evaluate this limit with or without applying derivatives? $$ \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} $$
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. 

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.
