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Question Number 124491 by Mammadli last updated on 03/Dec/20

lim_(n→∞) (2^n /(n!)) = 0

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}^{{n}} }{{n}!}\:=\:\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 03/Dec/20

let U_n =(2^n /(n!)) we have n! ∼ n^n e^(−n) (√(2πn)) ⇒ U_n ∼ (2^n /(n^n e^(−n) (√(2πn)))) =((2/n))^n e^n (2πn)^(−(1/2)) =e^(nln((2/n))+n) e^(−(1/2)ln(2πn)) =e^(nln((2/n))+n−(1/2)ln(2πn)) but lim_(n→+∞) nln((2/n))+n−(1/2)ln(2πn) =lim_(n→+∞) nln(2)−nln(n)+n−(1/2)ln(2πn) =lim_(n→+∞) nln(n){−1+((ln(2))/(ln(n)))−((ln(2πn))/(nln(n)))} =lim_(n→+∞) −nln(n)=−∞ ⇒lim_(n→+∞) e^((...)) =0 ⇒lim_(n→+∞) U_n =0

$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}!}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\:\sim\:\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} \sim\:\:\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}}\:=\left(\frac{\mathrm{2}}{\mathrm{n}}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{n}} \:\left(\mathrm{2}\pi\mathrm{n}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{e}^{\mathrm{nln}\left(\frac{\mathrm{2}}{\mathrm{n}}\right)+\mathrm{n}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right)} \:=\mathrm{e}^{\mathrm{nln}\left(\frac{\mathrm{2}}{\mathrm{n}}\right)+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right)} \\ $$$$\mathrm{but}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\mathrm{nln}\left(\frac{\mathrm{2}}{\mathrm{n}}\right)+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{nln}\left(\mathrm{2}\right)−\mathrm{nln}\left(\mathrm{n}\right)+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{nln}\left(\mathrm{n}\right)\left\{−\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{n}\right)}−\frac{\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{nln}\left(\mathrm{n}\right)}\right\}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} −\mathrm{nln}\left(\mathrm{n}\right)=−\infty\: \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{e}^{\left(...\right)} =\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\mathrm{0} \\ $$

Answered by mr W last updated on 04/Dec/20

(2^n /(n!))=((2×2×2×...×2)/(1×2×3×...×n))<((2/n))^n <(2/n) for n≥3 lim_(n→∞) (2^n /(n!))<lim_(n→∞) ((2/n))=0 ⇒lim_(n→∞) (2^n /(n!))=0

$$\frac{\mathrm{2}^{{n}} }{{n}!}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}×...×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×...×{n}}<\left(\frac{\mathrm{2}}{{n}}\right)^{{n}} <\frac{\mathrm{2}}{{n}}\:{for}\:{n}\geqslant\mathrm{3} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{n}} }{{n}!}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{{n}}\right)=\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{n}} }{{n}!}=\mathrm{0} \\ $$

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