Limit with sum of cube roots I am trying to evaluate
$$\lim \limits_{n \to \infty} \left(\sqrt[3]{{1 \over n^4}} + \sqrt[3]{{2 \over n^4}} + \sqrt[3]{{3 \over n^4}} + \dots + \sqrt[3]{{n \over n^4}} \right)$$
I simplify the expression to
$${\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n} \over \sqrt[3]{n^4}}$$
$y_n=\sqrt[3]{n^4} \to \infty$, so we might try to use the Stolz theorem to get the limit.
$${x_{n+1}-x_n \over y_{n+1}-y_n} = {\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{4} + \dots + \sqrt[3]{n+1} - \left(\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n}\right) \over \sqrt[3]{(n+1)^4}-\sqrt[3]{n^4}}={\sqrt[3]{n+1} \over \sqrt[3]{(n+1)^4} - \sqrt[3]{n^4}}$$
Whne I use the Stolz theorem, the limit seems to be $\infty$. Stolz works for $\infty$, so this would contradict the limit ${3 \over 4}$ obtained below by integration which I'm fairly confident is correct. Any suggestions on what I'm missing?

Hint:
$$
\sqrt[3]{a}-\sqrt[3]{b}=\frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{a\,b}+\sqrt[3]{b^2}}.
$$
Use it with both numerator and denominator in the expression you get from Stolz's criterium.
