Limit of sum with cube roots Do you have any hints for the following limit?
$$\lim\limits_{n \to \infty}{1 \over n^2}\left( {1 \over \sqrt[3]{1}} + {1 \over \sqrt[3]{2}} + {1 \over \sqrt[3]{3}} + \dots + {1 \over \sqrt[3]{n^2-1}} + {1 \over \sqrt[3]{n^2}} \right)$$

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 $n \to \infty$,
$$
\frac{1}{n^2} \sum_{k=1}^{n^2} \frac{1}{\sqrt[3]{k}}=\frac{1}{n^{2/3}} \cdot \frac{1}{n^2} \sum_{k=1}^{n^2} \sqrt[3]{\frac{n^2}{k}} \sim \frac{1}{n^{2/3}} \cdot \frac32 \to 0.
$$
