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Question Number 166926 by mathlove last updated on 02/Mar/22

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{5^n}{n!}=?$$

Answered by JDamian last updated on 02/Mar/22

$$0$$

Answered by Mathspace last updated on 02/Mar/22

$$u_n=\frac{5^n}{n!}\Rightarrow {logu}_n=nlog5-log(n!)$$$$n!\sim n^n{e}^{-n}\sqrt{2\pi n}\Rightarrow log(n!)$$$$=nlogn-n+\frac{1}{2}log\left(2\pi n\right)\Rightarrow$$$${logu}_n\sim nlog5-nlogn+n-\frac{1}{2}log\left(2\pi n\right)$$$$=n(log5-logn+1-\frac{1}{2}\frac{log\left(2\pi n\right)}{n})\to +\mathrm{\infty }$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{u}_n=0$$

Commented by mathlove last updated on 03/Mar/22

$$thanks$$

Answered by mr W last updated on 03/Mar/22

$$\frac{5^n}{n!}>\frac{1}{n!}$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{5^n}{n!}>\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n!}=0$$$$$$$$\frac{5^n}{n!}=\frac{5^{12}\times 5\times ..\times 5}{12!\times 13\times 14\times ...\times n}$$$$=\left(\frac{5^{12}}{12!}\right)\times \left(\frac{5}{13}\right)\times \left(\frac{5}{14}\right)\times ...\times \left(\frac{5}{n}\right)$$$$<\left(\frac{9765265}{19160964}\right)\times {\left(\frac{5}{13}\right)}^{n-12}$$$$<{\left(\frac{5}{13}\right)}^{n-12}$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{5^n}{n!}<\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{5}{13}\right)}^{n-12}=0$$$$0<\underset{n\to \mathrm{\infty }}{\lim }\frac{5^n}{n!}<0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\frac{5^n}{n!}=0$$

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