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$\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1}+2 \sqrt{2}+3 \sqrt{3}+\ldots+n \sqrt{n}}{n^{\frac{5}{2}}}\right)=$
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Verified Answer
The correct answer is:
$\frac{2}{5}$
We have,
$$
\lim _{x \rightarrow \infty}\left[\frac{\sqrt{1}+2 \sqrt{2}+3 \sqrt{3}+\ldots+n \sqrt{n}}{n^{5 / 2}}\right]
$$
$\begin{aligned} & =\lim _{x \rightarrow \infty} \sum_{r=1}^n\left(\frac{r}{n}\right)^{3 / 2} \cdot \frac{1}{n} \\ & =\int_0^1 x^{3 / 2} d x=\left[\frac{x^{5 / 2}}{5 / 2}\right]_0^1=\frac{2}{5}\end{aligned}$
$$
\lim _{x \rightarrow \infty}\left[\frac{\sqrt{1}+2 \sqrt{2}+3 \sqrt{3}+\ldots+n \sqrt{n}}{n^{5 / 2}}\right]
$$
$\begin{aligned} & =\lim _{x \rightarrow \infty} \sum_{r=1}^n\left(\frac{r}{n}\right)^{3 / 2} \cdot \frac{1}{n} \\ & =\int_0^1 x^{3 / 2} d x=\left[\frac{x^{5 / 2}}{5 / 2}\right]_0^1=\frac{2}{5}\end{aligned}$
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