Question

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

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

Answer 1

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

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

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{1/2}-\underset{n\to \infty }{lim}\frac{1}{n}\times {\left(\frac{n}{n}\right)}^{1/2}$
$=\underset{0}{\overset{1}{\int }}\sqrt{x}dx-0$
$={\left[\frac{x^{3/2}}{3/2}\right]}_0^1$
$=\frac{2}{3}$