Question

The value of $\underset{n\to \infty }{\lim }$$\left\{\frac{\sqrt{n+1}+\sqrt{n+2}+....+\sqrt{2n-1}}{n^{3/2}}\right\}$ is

• A. $\frac{2}{3}\left(2\sqrt{2}-1\right)$
• B. $\frac{2}{3}\left(\sqrt{2}-1\right)$
• C. $\frac{2}{3}\left(\sqrt{2}+1\right)$
• D. $\frac{2}{3}\left(2\sqrt{2}+1\right)$

Answer 1

Answer: (A)

$\underset{n\to \infty }{\lim }$$\left(\frac{\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+......+\sqrt{2n-1}}{n^{3/2}}\right)$
$=\underset{n\to \infty }{\lim }$$\left(\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{\sqrt{n}}}+.....+\sqrt{1+\frac{n-1}{n}}\right)\frac{1}{n}$
$=\underset{n\to \infty }{\lim }$ $\sum _{n=1}^{r-1}$$\frac{1}{n}\sqrt{1+\frac{r}{n}}$
$={\int }_0^1\sqrt{1+x}dx=\frac{2}{3}.\left(2\sqrt{2}-1\right)$