Question

$\underset{n\to \infty }{lim}\frac{\sqrt{1}+\sqrt{2}+.....+\sqrt{n-1}}{n\sqrt{n}}=0$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $0$ (zero)

Answer 1

The correct option is C $\frac{1}{3}$

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

$\underset{n\to \infty }{lim}\frac{\sqrt{1}+\sqrt{2}........\sqrt{n-1}}{n\sqrt{n}}=\underset{n\to \infty }{lim}\frac{{\int }_2^n\sqrt{x-1}dx}{n\sqrt{n}}=\underset{n\to \infty }{lim}[\frac{2}{3}\frac{(n-1{)}^{\frac{3}{2}}}{n\sqrt{n}}-\frac{1}{n\sqrt{n}}]=\underset{n\to \infty }{lim}\frac{2}{3}[{(1-\frac{1}{n})}^{\frac{3}{2}}-\frac{1}{n^{\frac{3}{2}}}]=\frac{2}{3}[1-0]=\frac{2}{3}$

Hence the correct answer is $\frac{2}{3}$