limn→∞ n!nn1/n is equal to

Let,P=limn→∞ n!nn1/n=limn→∞ 1n⋅2n⋅3n⋅4n⋅nn1/n∴log⁡P=limn→∞ 1nlog⁡1n+log⁡2n+…+log⁡nn⇒log⁡P=limn→∞ ∑r=1n 1nlog⁡rn⇒log⁡P=∫01 log⁡xdx=[xlog⁡x−x]01=−1⇒P=1e

limn→∞ n!nn1/n is equal to

Let,P=limn→∞ n!nn1/n=limn→∞ 1n⋅2n⋅3n⋅4n⋅nn1/n∴log⁡P=limn→∞ 1nlog⁡1n+log⁡2n+…+log⁡nn⇒log⁡P=limn→∞ ∑r=1n 1nlog⁡rn⇒log⁡P=∫01 log⁡xdx=[xlog⁡x−x]01=−1⇒P=1e

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$lim_{n \to \infty} {[\frac{n!}{n^{n}}]}^{1 / n}$ is equal to

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1/e

$\frac{π}{4}$

$\frac{4}{π}$

answer is B.

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Detailed Solution

Let,

$\begin{array}{l} P = lim_{n \to \infty} {[\frac{n!}{n^{n}}]}^{1 / n} = lim_{n \to \infty} {[\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot \frac{4}{n} \cdot \frac{n}{n}]}^{1 / n} \\ ∴ \log P = lim_{n \to \infty} \frac{1}{n} [\log \frac{1}{n} + \log \frac{2}{n} + \dots + \log \frac{n}{n}] \\ \Rightarrow \log P = lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \log \frac{r}{n} \\ \Rightarrow \log P = \int_{0}^{1} \log xdx = [xlog x - x]_{0}^{1} = - 1 \Rightarrow P = \frac{1}{e} \end{array}$