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Question: If \(\lim_{n \rightarrow \infty}\left\lbrack \frac{n!}{n^{n}} \right\rbrack^{1/n}\) equal...

If lim⁡n→∞[n!nn]1/n\lim_{n \rightarrow \infty}\left\lbrack \frac{n!}{n^{n}} \right\rbrack^{1/n} equal

A

e

B

1e\frac{1}{e}

C

π4\frac{\pi}{4}

D

4π\frac{4}{\pi}

Answer

1e\frac{1}{e}

Explanation

Solution

Let P=lim⁡n→∞(n!nn)1/n⇒P=lim⁡n→∞(1n.2n.3n.4n..........nn)1/nP = \lim_{n \rightarrow \infty}\left( \frac{n!}{n^{n}} \right)^{1/n} \Rightarrow P = \lim_{n \rightarrow \infty}\left( \frac{1}{n}.\frac{2}{n}.\frac{3}{n}.\frac{4}{n}..........\frac{n}{n} \right)^{1/n}

∴log⁡P=1nlim⁡n→∞(log⁡1n+log⁡2n+.........+log⁡nn)\therefore\log P = \frac{1}{n}\lim_{n \rightarrow \infty}\left( \log\frac{1}{n} + \log\frac{2}{n} + ......... + \log\frac{n}{n} \right)

⇒\Rightarrow log⁡P=lim⁡n→∞∑r=1n1nlog⁡rn\log P = \lim _ { n \rightarrow \infty } \sum _ { r = 1 } ^ { n } \frac { 1 } { n } \log \frac { r } { n } log⁡P=∫01log⁡xdx=[xlog⁡x−x]01⥂=(−1)⇒P=1e\log P = \int_{0}^{1}{\log xdx = \lbrack x\log x - x\rbrack_{0}^{1} ⥂ = ( - 1)} \Rightarrow P = \frac{1}{e}.