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Question Number 198855 by mokys last updated on 25/Oct/23
prove : lim_(n→∞) ((n!)/(n^x (n−x)!)) = 1
$${prove}\::\:\underset{{n}\rightarrow\infty} {{lim}}\:\frac{{n}!}{{n}^{{x}} \left({n}−{x}\right)!}\:=\:\mathrm{1}\: \\ $$
Answered by witcher3 last updated on 25/Oct/23
((Γ(n+1))/(Γ(n+1−x)n^x )) Γ(z)=(√(2π)).z^(z−(1/2)) e^(−z) ((Γ(n+1))/(n^x (n−x)!))∼(((√(2π))(n+1)^(n+(1/2)) e^(−(n+1)) )/(n^x (√(2π)).(n+1−x)^(n+(1/2)−x) e^(−(n+1−x)) )) =lim_(n→∞) .(e^(−x) /n^x ).(((n+1)/(n+1−x)))^(n+(1/2)) .(1/((n+1−x)^(−x) )) (((n+1)/(n+1−x)))^(n+(1/2)) =e^((n+(1/2))ln(1+(x/(n+1−x)))) =e^((n+(1/2)).((x/(n+1−x))−(1/2)((x/(n+1−x)))^2 +o((1/n^2 ))) ∼e^x n^x .(n+1−x)^(−x) =((n/(n+1−x)))^x =lim_(n→∞) e^(xln(1+((x−1)/(n+1−x)))) →1 (e^(−x) /n^x ).(((n+1)/(n+1−x)))^(n+(1/2)) .(1/((n+1−x)^(−x) )) =(((n+1)/(n+1−x)))^(n+(1/2)) .e^(−x) .((n/(n+1−x)))^(−x) ∼e^x .e^(−x) .1∼1 ⇒lim_(n→∞) .((Γ(n+1))/(Γ(n+1−x).n^x ))=1
$$\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)\mathrm{n}^{\mathrm{x}} } \\ $$$$\Gamma\left(\mathrm{z}\right)=\sqrt{\mathrm{2}\pi}.\mathrm{z}^{\mathrm{z}−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{z}} \\ $$$$\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}^{\mathrm{x}} \left(\mathrm{n}−\mathrm{x}\right)!}\sim\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)} }{\mathrm{n}^{\mathrm{x}} \sqrt{\mathrm{2}\pi}.\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{x}} \mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)} } \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{n}^{\mathrm{x}} }.\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)^{−\mathrm{x}} } \\ $$$$\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{e}^{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)} \\ $$$$=\mathrm{e}^{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right).\left(\frac{\mathrm{x}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{x}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right.} \\ $$$$\sim\mathrm{e}^{\mathrm{x}} \\ $$$$\mathrm{n}^{\mathrm{x}} .\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)^{−\mathrm{x}} =\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{x}} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}e}^{\mathrm{xln}\left(\mathrm{1}+\frac{\mathrm{x}−\mathrm{1}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)} \rightarrow\mathrm{1} \\ $$$$\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{n}^{\mathrm{x}} }.\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)^{−\mathrm{x}} } \\ $$$$=\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} .\mathrm{e}^{−\mathrm{x}} .\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}−\mathrm{x}}\overset{−\mathrm{x}} {\right)}\sim\mathrm{e}^{\mathrm{x}} .\mathrm{e}^{−\mathrm{x}} .\mathrm{1}\sim\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right).\mathrm{n}^{\mathrm{x}} }=\mathrm{1} \\ $$$$ \\ $$
Commented by mokys last updated on 25/Oct/23
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by witcher3 last updated on 27/Oct/23
Withe Pleasur
$$\mathrm{Withe}\:\mathrm{Pleasur} \\ $$

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