The value of lim _{n rightarrow infty} left{ | vedclass

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(A) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^{3/2}} \sum_{r=1}^{n-1} \sqrt{n+r}$ Dividing the numerator and denominator by $n$, we ge

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$\frac{2}{3}(2\sqrt{2}-1)$

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(A) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^{3/2}} \sum_{r=1}^{n-1} \sqrt{n+r}$ Dividing the numerator and denominator by $n$, we get: $= \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{\frac{n+r}{n}} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{1+\frac{r}{n}}$ This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $f(x) = \sqrt{1+x}$ $= \int_{0}^{1} (1+x)^{1/2} dx = \left[ \frac{(1+x)^{3/2}}{3/2} \right]_{0}^{1} = \frac{2}{3} \left[ (1+x)^{3/2} \right]_{0}^{1}$ $= \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$

The value of $\lim _{n \rightarrow \infty} \left\{ \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2n-1}}{n^{3/2}} \right\}$ is

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