For natural number n, $(n!)^{2}$ > $n^{n}$ , if

n > 3

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n > 4

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$n \geq$ 4

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$n \geq$ 3

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Solution

The correct option is D
$n \geq$ 3

Check through option, condition $(n!)^{2}$ > $n^{n}$ is

true when n ≥ 3.

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Similar questions

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

(n!)^{2} > n^{n}

Statement I

n \geq 2

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement II

n \geq 2

\sqrt{n (n + 1)} < n + 1

2^{n} (n - 1)! < n^{n}

n \geq 2, \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

n \geq 2, \sqrt{n (n + 1)} < n + 1

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