Question

The value of $\underset{n\to \infty }{lim}\left\{\begin{array}{c}\frac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\frac{3}{2}}}\end{array}\right\}$ 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 $\frac{2}{3}(2\sqrt{2}-1)$

$\underset{n\to \infty }{lim}\left\{\begin{array}{c}\frac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\frac{3}{2}}}\end{array}\right\}$

$=\underset{n\to \infty }{lim}\left\{\begin{array}{c}\frac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{\sqrt{n}}\end{array}\right\}\frac{1}{n}$

$=\underset{n\to \infty }{lim}\left\{\begin{array}{c}\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+.....+\sqrt{1+\frac{n-1}{n}}\end{array}\right\}\frac{1}{n}$

Now using limit as an integrals

$={\int }_0^1\sqrt{1+x}dx=\frac{2}{3}\left[\right(1+x)\sqrt{1+x}{]}_0^1=\frac{2}{3}(2\sqrt{2}-1)$