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The value of $\lim _{n \rightarrow \infty}\left\{\frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2 n-1}}{n^{3 / 2}}\right\}$ is
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$\frac{2}{3}(2 \sqrt{2}-1)$
$\because \lim _{n \rightarrow \infty \left[\frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2 n-1}}{n^{\frac{3}{2}}}\right]}$
$=\lim _{n \rightarrow \infty 1}\left[\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+\ldots+\sqrt{1+\frac{n-1}{n}}\right] \frac{1}{n}$
$=\lim _{n \rightarrow \infty 1} \sum_{r=1}^{n-1} \frac{1}{n} \sqrt{1+\frac{r}{n}}$
$=\int_{0}^{1} \sqrt{1+x} d x=\left[\frac{2}{3}(1+x)^{3 / 2}\right]_{0}^{1}=\frac{2}{3}(2 \sqrt{2}-1)$
$=\lim _{n \rightarrow \infty 1}\left[\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+\ldots+\sqrt{1+\frac{n-1}{n}}\right] \frac{1}{n}$
$=\lim _{n \rightarrow \infty 1} \sum_{r=1}^{n-1} \frac{1}{n} \sqrt{1+\frac{r}{n}}$
$=\int_{0}^{1} \sqrt{1+x} d x=\left[\frac{2}{3}(1+x)^{3 / 2}\right]_{0}^{1}=\frac{2}{3}(2 \sqrt{2}-1)$
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