Limit of sum with cube roots

Question

Do you have any hints for the following limit?

$$\lim\limits_{n \to \infty}{1 \over n^2}\left( {1 \over \sqrt[3]{1}} + {1 \over \sqrt[3]{2}} + {1 \over \sqrt[3]{3}} + \dots + {1 \over \sqrt[3]{n^2-1}} + {1 \over \sqrt[3]{n^2}} \right)$$

Answer 1

hint:Let $a_k = \dfrac{1}{\sqrt[3]{k}}\to 0$. Cesaro's theorem says: $S_n = \dfrac{a_1+a_2+\cdots + a_n}{n} \to 0$, thus the subsequence $S_{n^2} \to 0$ as well.

Answer 2

One may use a Riemann sum, as $n \to \infty$, $$ \frac{1}{n^2} \sum_{k=1}^{n^2} \sqrt[3]{\frac{n^2}{k}}\to \int_0^1 \frac{1}{\sqrt[3]{x}}\:dx=\frac32 $$ giving, as $n \to \infty$, $$ \frac{1}{n^2} \sum_{k=1}^{n^2} \frac{1}{\sqrt[3]{k}}=\frac{1}{n^{2/3}} \cdot \frac{1}{n^2} \sum_{k=1}^{n^2} \sqrt[3]{\frac{n^2}{k}} \sim \frac{1}{n^{2/3}} \cdot \frac32 \to 0. $$