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

lim_(n→∞) (5^n /(n!))=?

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

Answered by JDamian last updated on 02/Mar/22

0

$$\mathrm{0} \\ $$

Answered by Mathspace last updated on 02/Mar/22

u_n =(5^n /(n!)) ⇒logu_n =nlog5−log(n!)  n!∼n^n e^(−n) (√(2πn))⇒log(n!)  =nlogn−n+(1/2)log(2πn) ⇒  logu_n ∼nlog5−nlogn+n−(1/2)log(2πn)  =n(log5−logn+1−(1/2)((log(2πn))/n))→+∞  ⇒lim_(n→+∞) u_n =0

$${u}_{{n}} =\frac{\mathrm{5}^{{n}} }{{n}!}\:\Rightarrow{logu}_{{n}} ={nlog}\mathrm{5}−{log}\left({n}!\right) \\ $$$${n}!\sim{n}^{{n}} {e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\Rightarrow{log}\left({n}!\right) \\ $$$$={nlogn}−{n}+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\pi{n}\right)\:\Rightarrow \\ $$$${logu}_{{n}} \sim{nlog}\mathrm{5}−{nlogn}+{n}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\pi{n}\right) \\ $$$$={n}\left({log}\mathrm{5}−{logn}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\frac{{log}\left(\mathrm{2}\pi{n}\right)}{{n}}\right)\rightarrow+\infty \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =\mathrm{0} \\ $$

Commented by mathlove last updated on 03/Mar/22

thanks

$${thanks} \\ $$

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

(5^n /(n!))>(1/(n!))  ⇒lim_(n→∞) (5^n /(n!))>lim_(n→∞) (1/(n!))=0    (5^n /(n!))=((5^(12) ×5×..×5)/(12!×13×14×...×n))  =((5^(12) /(12!)))×((5/(13)))×((5/(14)))×...×((5/n))  <(((9 765 265)/(19 160 964)))×((5/(13)))^(n−12)   <((5/(13)))^(n−12)   ⇒lim_(n→∞) (5^n /(n!))<lim_(n→∞) ((5/(13)))^(n−12) =0  0<lim_(n→∞) (5^n /(n!)) <0  ⇒lim_(n→∞) (5^n /(n!)) =0

$$\frac{\mathrm{5}^{{n}} }{{n}!}>\frac{\mathrm{1}}{{n}!} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}^{{n}} }{{n}!}>\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}!}=\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{5}^{{n}} }{{n}!}=\frac{\mathrm{5}^{\mathrm{12}} ×\mathrm{5}×..×\mathrm{5}}{\mathrm{12}!×\mathrm{13}×\mathrm{14}×...×{n}} \\ $$$$=\left(\frac{\mathrm{5}^{\mathrm{12}} }{\mathrm{12}!}\right)×\left(\frac{\mathrm{5}}{\mathrm{13}}\right)×\left(\frac{\mathrm{5}}{\mathrm{14}}\right)×...×\left(\frac{\mathrm{5}}{{n}}\right) \\ $$$$<\left(\frac{\mathrm{9}\:\mathrm{765}\:\mathrm{265}}{\mathrm{19}\:\mathrm{160}\:\mathrm{964}}\right)×\left(\frac{\mathrm{5}}{\mathrm{13}}\right)^{{n}−\mathrm{12}} \\ $$$$<\left(\frac{\mathrm{5}}{\mathrm{13}}\right)^{{n}−\mathrm{12}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}^{{n}} }{{n}!}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5}}{\mathrm{13}}\right)^{{n}−\mathrm{12}} =\mathrm{0} \\ $$$$\mathrm{0}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}^{{n}} }{{n}!}\:<\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}^{{n}} }{{n}!}\:=\mathrm{0} \\ $$

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