The correct option is C 23n →∞lim√(1) + √(2) + ... + √(n 1)n√(n) =n →∞lim1n∑_r=1^n( rn)^1/2 n →∞lim nbsp;1/n×(n/n)^1/2 = nbsp;∫_0^1√(x) dx 0 nbsp; = ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

lim n →∞√1+√2...

Question

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

\frac{1}{2}

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\frac{1}{3}

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\frac{2}{3}

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0

(zero)

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Solution

The correct option is C $\frac{2}{3}$
$lim n \to \infty \frac{\sqrt{1} + \sqrt{2} + . . . + \sqrt{n - 1}}{n \sqrt{n}}$

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

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Similar questions

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Q. Evalute

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

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is equal to

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limn→∞√1+√2+...+√n−1n√n=

The correct option is C 23limn→∞√1+√2+...+√n−1n√n =limn→∞1nn∑r=1(rn)1/2−limn→∞1n×(nn)1/2 =1∫0√x dx−0 =[x3/23/2]10 =23

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