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Question Number 95363 by john santu last updated on 24/May/20
lim_(x→∞) (((x!)^2 )/((2x)!)) = ?
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}!\right)^{\mathrm{2}} }{\left(\mathrm{2x}\right)!}\:=\:? \\ $$
Commented by john santu last updated on 24/May/20
Answered by mr W last updated on 24/May/20
(((x!)^2 )/((2x)!))=(((x!)(x!))/((x!)(x+1)(x+2)...(x+x))) =((1∙2∙....∙x)/((x+1)(x+2)...(x+x))) =(1/(((x/1)+1)((x/2)+1)...((x/x)+1)))<(1/(((x/x)+1)^x ))=(1/2^x ) lim_(x→∞) (((x!)^2 )/((2x)!)) <lim_(x→∞) (1/2^x )=0 ⇒lim_(x→∞) (((x!)^2 )/((2x)!)) =0
$$\frac{\left({x}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{x}\right)!}=\frac{\left({x}!\right)\left({x}!\right)}{\left({x}!\right)\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{x}\right)} \\ $$$$=\frac{\mathrm{1}\centerdot\mathrm{2}\centerdot….\centerdot{x}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\frac{{x}}{\mathrm{1}}+\mathrm{1}\right)\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)…\left(\frac{{x}}{{x}}+\mathrm{1}\right)}<\frac{\mathrm{1}}{\left(\frac{{x}}{{x}}+\mathrm{1}\right)^{{x}} }=\frac{\mathrm{1}}{\mathrm{2}^{{x}} } \\ $$$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}!\right)^{\mathrm{2}} }{\left(\mathrm{2x}\right)!}\:<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}^{{x}} }=\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}!\right)^{\mathrm{2}} }{\left(\mathrm{2x}\right)!}\:=\mathrm{0} \\ $$
Commented by john santu last updated on 24/May/20
by sequeze theorem?
$$\mathrm{by}\:\mathrm{sequeze}\:\mathrm{theorem}? \\ $$
Answered by mathmax by abdo last updated on 24/May/20
we have n! ∼n^n e^(−n) (√(2πn))⇒(n!)^2 ∼n^(2n) e^(−2n) (2πn) (2n)! ∼(2n)^(2n) e^(−2n) (√(2π(2n))) =4^n n^(2n) e^(−2n) 2(√(πn)) ⇒ (((n!)^2 )/((2n)!)) ∼((n^(2n) e^(−2n) (2πn))/(4^n n^(2n) e^(−2n) (2(√(πn))))) =((√(nπ))/4^n ) =(√π)n^(1/2) .e^(−nln(4)) →0(n→∞) ⇒lim_(n→+∞) (((n!)^2 )/((2n)!)) =0
$$\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\:\sim\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\Rightarrow\left(\mathrm{n}!\right)^{\mathrm{2}} \sim\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{n}\right) \\ $$$$\left(\mathrm{2n}\right)!\:\sim\left(\mathrm{2n}\right)^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \sqrt{\mathrm{2}\pi\left(\mathrm{2n}\right)}\:=\mathrm{4}^{\mathrm{n}} \:\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \mathrm{2}\sqrt{\pi\mathrm{n}}\:\Rightarrow \\ $$$$\frac{\left(\mathrm{n}!\right)^{\mathrm{2}} }{\left(\mathrm{2n}\right)!}\:\sim\frac{\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{4}^{\mathrm{n}} \:\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\sqrt{\pi\mathrm{n}}\right)}\:=\frac{\sqrt{\mathrm{n}\pi}}{\mathrm{4}^{\mathrm{n}} }\:=\sqrt{\pi}\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{2}}} \:.\mathrm{e}^{−\mathrm{nln}\left(\mathrm{4}\right)} \:\rightarrow\mathrm{0}\left(\mathrm{n}\rightarrow\infty\right) \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\left(\mathrm{n}!\right)^{\mathrm{2}} }{\left(\mathrm{2n}\right)!}\:=\mathrm{0} \\ $$

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