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Question Number 186323 by Rupesh123 last updated on 03/Feb/23

Commented by Frix last updated on 03/Feb/23

$$n!\sim \frac{n^n}{{\mathrm{e}}^n}\sqrt{2\pi n}$$$$a_n\sim \frac{\sqrt{2\pi n}}{{\mathrm{e}}^n}$$$$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=0$$

Answered by JDamian last updated on 03/Feb/23

$$\underset{―}{\mathrm{just}\mathrm{a}\mathrm{bit}\mathrm{of}\mathrm{thinking}}$$$$\mathrm{both}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{are}$$$$\mathrm{products}\mathrm{of}\mathit{n}\mathrm{factors}$$$$a_n=\frac{\cancel{n}}{\cancel{n}}\cdot \frac{n-1}{n}\cdot \frac{n-2}{n}\cdot \cdot \cdot \frac{2}{n}\cdot \frac{1}{n}$$$$\mathrm{every}\mathrm{fraction}\mathrm{is}<1$$$$a_n<1\mathrm{\forall }n>1$$$$\mathrm{Denominator}\mathrm{grows}\mathrm{faster}\mathrm{than}\mathrm{numerator}$$$$\mathrm{because}{\mathrm{n}}^{\mathrm{n}}>\mathrm{n}!\mathrm{\forall }n>1$$

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