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$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to
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$\frac{2}{3}$
$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$
$=\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}+\sqrt{n}}{n \sqrt{n}}-\frac{n}{n \times n}\right)$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n}\left(\frac{r}{n}\right)^{1 / 2}-\lim _{n \rightarrow \infty} \frac{1}{n} \times \frac{n}{n}=\int_{0}^{1} \sqrt{x} d x+0=\frac{2}{3}$
$=\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}+\sqrt{n}}{n \sqrt{n}}-\frac{n}{n \times n}\right)$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n}\left(\frac{r}{n}\right)^{1 / 2}-\lim _{n \rightarrow \infty} \frac{1}{n} \times \frac{n}{n}=\int_{0}^{1} \sqrt{x} d x+0=\frac{2}{3}$
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