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?

Let $f(x)=x^\frac 13$.
$$S_n=\sum_{k=1}^n (\frac{k}{n^4})^{\frac 13}\;\;$$
is a Riemann sum and
$$\lim_{n\to+\infty}S_n=$$
$$\lim_{n\to+\infty}\frac{1-0}{n} \sum_{k=1}^n f(0+k\frac{1-0}{n})$$
$$=\int_0^1 x^\frac{1}{3}dx$$
$$=\frac{3}{4}$$
