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

Answer 1

The correct option is A $\frac{2}{3}(2\sqrt{2}-1)$

$\underset{n\to \infty }{lim}\left\{\frac{\sqrt{n+1}+\sqrt{n+2}+...+\sqrt{2n-1}}{n^{3/2}}\right\}=\underset{n\to \infty }{lim}\frac{1}{n}\left\{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+...+\sqrt{1+\frac{n-1}{n}}\right\}=\underset{n\to \infty }{lim}\sum _{r=1}^{n-1}\frac{1}{n}\sqrt{1+\frac{r}{n}}=\underset{0}{\overset{1}{\int }}\sqrt{1+x}dx=\frac{2}{3}(2\sqrt{2}-1)$