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Question Number 30759 by abdo imad last updated on 25/Feb/18

$$find{lim}_{n\to \mathrm{\infty }}{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}.$$

Commented by abdo imad last updated on 27/Feb/18

$$letusethestirlingformulawehave$$$$n!\sim n^n{e}^{-n}\sqrt{2\pi n}\Rightarrow \frac{n!}{n^n}\sim e^{-n}\sqrt{2\pi n}\Rightarrow$$$${\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}={\left(e^{-n}\sqrt{2\pi n}\right)}^{\frac{1}{n}}=e^{-1}{\left(2\pi n\right)}^{\frac{1}{2n}}but$$$${\left(2\pi n\right)}^{\frac{1}{2n}}=e^{\frac{1}{2n}ln\left(2\pi n\right)}=e^{\frac{ln\left(2\pi \right)}{2n}+\frac{ln\left(n\right)}{2n}}\to 1(n\to \mathrm{\infty })\Rightarrow$$$${lim}_{n\to \mathrm{\infty }}{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}=\frac{1}{e}.$$

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