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Question Number 106505 by Ar Brandon last updated on 05/Aug/20

Show thatlim_(n→∞) [((n!)/((n−x)!n^x ))]=1

$$\mathrm{Show}\:\mathrm{that}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{x}\right)!\mathrm{n}^{\mathrm{x}} }\right]=\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 05/Aug/20

lim_(n→∞) ((n(n−1)....x times×(n−x)!)/((n−x)!n^x )) lim_(n→∞) ((n(n−1)....xtimes)/(n.n....x times)) lim_(n→∞) (1−(1/n))(1−(2/n))....=1

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}\left({n}−\mathrm{1}\right)....{x}\:{times}×\left({n}−{x}\right)!}{\left({n}−{x}\right)!{n}^{{x}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}\left({n}−\mathrm{1}\right)....{xtimes}}{{n}.{n}....{x}\:{times}} \\ $$$$\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)....=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 05/Aug/20

��Thanks. �� I arrived there while trying to demonstrate that as n and p go to infinity and zero respectively, the binomial probability distribution becomes approximately equal to the poisson distribution.

Commented by Dwaipayan Shikari last updated on 05/Aug/20

������

Commented by Dwaipayan Shikari last updated on 05/Aug/20

Are you from France?

Commented by Dwaipayan Shikari last updated on 05/Aug/20

Là nuit Bonne nuit

Commented by mathmax by abdo last updated on 06/Aug/20

here x is real not integr ...!

$$\mathrm{here}\:\mathrm{x}\:\mathrm{is}\:\mathrm{real}\:\mathrm{not}\:\mathrm{integr}\:...! \\ $$

Commented by Ar Brandon last updated on 06/Aug/20

����Bonne Nuit, mon ami Shikari.��

Answered by mathmax by abdo last updated on 06/Aug/20

n! ∼ n^n e^(−n) (√(2πn))(stirling) (n−x)! ∼ (n−x)^(n−x) e^(−(n−x)) (√(2π(n−x))) ⇒ (n−x)! n^x ∼ (n−x)^n (n−x)^(−x) n^x e^(−(n−x)) (√(2π(n−x))) =(n−x)^n ((n/(n−x)))^x e^(−n) e^x (√(2π(n−x))) ⇒ ((n!)/((n−x)!n^x )) ∼ ((n^n e^(−n) (√(2πn)))/((n−x)^n ((n/(n−x)))^x e^(−n) e^x (√(2π(n−x))))) =((n/(n−x)))^n .(((n−x)/n))^(−x) e^(−x) (√((2πn)/(2π(n−x))))but (((n−x)/n))^x =(1−(x/n))^x →1 and ((n/(n−x)))^n =(((n−x)/n))^(−x) =(1−(x/n))^(−n) =e^(−nln(1−(x/n))) →e^x and (√(n/(n−x)))→1 ⇒ lim_(n→+∞) ((n!)/((n−x)!n^x )) =1

$$\mathrm{n}!\:\sim\:\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\left(\mathrm{stirling}\right) \\ $$$$\left.\left(\mathrm{n}−\mathrm{x}\right)!\:\sim\:\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{n}−\mathrm{x}} \:\mathrm{e}^{−\left(\mathrm{n}−\mathrm{x}\right)} \sqrt{\mathrm{2}\pi\left(\mathrm{n}−\mathrm{x}\right.}\right)\:\Rightarrow \\ $$$$\left(\mathrm{n}−\mathrm{x}\right)!\:\mathrm{n}^{\mathrm{x}} \:\sim\:\:\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{x}\right)^{−\mathrm{x}} \:\mathrm{n}^{\mathrm{x}} \:\:\mathrm{e}^{−\left(\mathrm{n}−\mathrm{x}\right)} \sqrt{\mathrm{2}\pi\left(\mathrm{n}−\mathrm{x}\right)}\: \\ $$$$=\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{n}} \left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{x}}\right)^{\mathrm{x}} \:\mathrm{e}^{−\mathrm{n}} \:\mathrm{e}^{\mathrm{x}} \sqrt{\mathrm{2}\pi\left(\mathrm{n}−\mathrm{x}\right)}\:\Rightarrow \\ $$$$\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{x}\right)!\mathrm{n}^{\mathrm{x}} }\:\sim\:\frac{\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}}{\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{n}} \:\left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{x}}\right)^{\mathrm{x}} \:\mathrm{e}^{−\mathrm{n}} \:\mathrm{e}^{\mathrm{x}} \sqrt{\mathrm{2}\pi\left(\mathrm{n}−\mathrm{x}\right)}} \\ $$$$=\left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{x}}\right)^{\mathrm{n}} .\left(\frac{\mathrm{n}−\mathrm{x}}{\mathrm{n}}\right)^{−\mathrm{x}} \:\:\mathrm{e}^{−\mathrm{x}} \sqrt{\frac{\mathrm{2}\pi\mathrm{n}}{\mathrm{2}\pi\left(\mathrm{n}−\mathrm{x}\right)}}\mathrm{but} \\ $$$$\left(\frac{\mathrm{n}−\mathrm{x}}{\mathrm{n}}\right)^{\mathrm{x}} =\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{n}}\right)^{\mathrm{x}} \:\rightarrow\mathrm{1}\:\:\mathrm{and}\:\:\left(\frac{\mathrm{n}}{\mathrm{n}−\mathrm{x}}\right)^{\mathrm{n}} \:=\left(\frac{\mathrm{n}−\mathrm{x}}{\mathrm{n}}\right)^{−\mathrm{x}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{n}}\right)^{−\mathrm{n}} =\mathrm{e}^{−\mathrm{nln}\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{n}}\right)} \:\rightarrow\mathrm{e}^{\mathrm{x}} \:\:\mathrm{and}\:\sqrt{\frac{\mathrm{n}}{\mathrm{n}−\mathrm{x}}}\rightarrow\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\:\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{x}\right)!\mathrm{n}^{\mathrm{x}} }\:=\mathrm{1} \\ $$$$\:\:\:\:\:\: \\ $$

Commented by Ar Brandon last updated on 06/Aug/20

Thanks Sir. Sorry I didn't precise that. In my case x is a natural number. But still arrived at thesame result. Thanks��

Commented by Dwaipayan Shikari last updated on 06/Aug/20

Thanking you for your corrections

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