limn→∞√1+√2+…+√nn^32=

Correct answer: (B) 2/3. The expression is given as, I=limn→∞(√1+√2+….+√nn^32=limn→∞√1+√2+…..+√n)/(n√n) =limn→∞1/n√1n+√2n+…+√nn =∑r=1nlimn→∞1/n√1n =∫01√xdx =2/3x^3201=2/3

Q116 / 160 · AP EAMCET 2019 (21 Apr Shift 2) · Mathematics

$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$

A
$0$
Correct answer
B
$\frac{2}{3}$
C
$1$
D
$\frac{3}{2}$

Solution

The expression is given as,

$I = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots . + \sqrt{n}}{n^{\frac{3}{2}}}$

$= \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots . . + \sqrt{n}}{n \sqrt{n}}$

$= \lim_{n \to \infty} \frac{1}{n} [\sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \dots + \sqrt{\frac{n}{n}}]$

$= \sum_{r = 1}^{n} \lim_{n \to \infty} \frac{1}{n} \sqrt{\frac{1}{n}}$

$= \int_{0}^{1} \sqrt{x} d x$

$= \frac{2}{3} {[x^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3}$

Hence, the correct answer is option: 'B'.

limn→∞1+2+…+nn32=

Solution

Relevant Questions

$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$