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Question Number 29509 by abdo imad last updated on 09/Feb/18

$$find{lim}_{n\to +\mathrm{\infty }}\frac{n!}{2^{n-1}}.$$

Commented by abdo imad last updated on 11/Feb/18

$$letusestirlingformulan!\sim n^n{e}^{-n}\sqrt{2\pi n}\Rightarrow$$$$\frac{n!}{2^{n-1}}=n^n{2}^{1-n}{e}^{-n}{n}^{\frac{1}{2}}\sqrt{2\pi }=e^{nln\left(n\right)}{e}^{(1-n)ln\left(2\right)-n+\frac{1}{2}}\sqrt{2\pi }$$$$=\sqrt{2\pi }{e}^{nln\left(n\right)+(1-n)ln\left(2\right)-n+\frac{1}{2}}=\sqrt{2\pi }{e}^{n(lnn+(\frac{1}{n}-1)ln2-1+\frac{1}{2n})}$$$$andlim\frac{n!}{2^{n-1}}={lim}_{n\to \mathrm{\infty }}\sqrt{2\pi }{e}^{n(ln\left(n\right)-1)}=+\mathrm{\infty }.$$

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