Another limit of summation Please help - how to solve this:

$$\lim_{n \to \infty}\frac1{n}\sum_{k=0}^{n}{\frac{1}{1+\frac{k^6}{n^3}}} $$


It seems the following.
Put $$f(k,n)= \frac{1}{1+\frac{k^6}{n^3}}.$$ Then $$\sum_{k=0}^{n} f(k,n)\le \sum_{k\le n^{2/3}}f(k,n) + \sum_{k\ge n^{2/3}} f(k,n)
\le n^{2/3}+\frac{n}{1+n}.$$ Hence $$0\le \lim_{n \to \infty}\frac1{n}\sum_{k=0}^{n} f(k,n)\le \lim_{n \to \infty} \frac1{n}\left(n^{2/3}+\frac{n}{1+n}\right)=0.$$
