Question

$\underset{n\to \infty }{lim}\frac{\sqrt{1}+\sqrt{2}+...+\sqrt{n-1}}{n\sqrt{n}}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $0$

Answer 1

The correct option is C $\frac{2}{3}$

$\underset{n\to \infty }{lim}\frac{\sqrt{1}+\sqrt{2}+...+\sqrt{n-1}}{n\sqrt{n}}$

$=\underset{n\to \infty }{lim}(\frac{\sqrt{1}+\sqrt{2}+...+\sqrt{n-1}+\sqrt{n}}{n\sqrt{n}}-\frac{1}{n})$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{1/2}-\underset{n\to \infty }{lim}\frac{1}{n}\times \frac{n}{n}$

$={\int }_0^1\sqrt{x}dx+0=\frac{2}{3}$