find-lim-n-n-2-n-1-

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

$${find}\:{lim}_{{n}\rightarrow+\:\infty} \:\:\:\:\frac{{n}!}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$

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

let use stirling formula n! ∼ n^n e^(−n) (√(2πn)) ⇒ ((n!)/2^(n−1) )= n^n 2^(1−n) e^(−n) n^(1/2) (√(2π)) = e^(nln(n)) e^((1−n)ln(2)−n+(1/2)) (√(2π)) = (√(2π)) e^(nln(n)+(1−n)ln(2)−n+(1/2)) = (√(2π)) e^(n(lnn +((1/n)−1)ln2−1 +(1/(2n)))) and lim ((n!)/2^(n−1) )=lim_(n→∞) (√(2π)) e^(n(ln(n)−1)) =+∞ .

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

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