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Question Number 106591 by mathdave last updated on 06/Aug/20

Commented by Her_Majesty last updated on 06/Aug/20

$$\mathrm{\infty }isnotanumber$$$$\underset{x\to \mathrm{\infty }}{limx}!\ne \sqrt{2\pi }$$

Commented by mathdave last updated on 06/Aug/20

$$yesbutcanbprove$$

Commented by Her_Majesty last updated on 06/Aug/20

$${it}^{\prime }ssimplywrong.$$$$x!\approx {\left(\frac{x}{\mathrm{e}}\right)}^x\sqrt{2\pi x}forx\to +\mathrm{\infty }wegetx!\to +\mathrm{\infty }$$

Commented by Her_Majesty last updated on 07/Aug/20

$$I^{\prime }mnotaSir\mathrm{beware}!!!I^{\prime }maLady$$

Commented by Ar Brandon last updated on 07/Aug/20

Lol, don't bother yourself Sir. I know you wouldn't like to draw the others' attention��

Commented by Her_Majesty last updated on 07/Aug/20

$$whatareyoutalkingabout?$$

Commented by Ar Brandon last updated on 07/Aug/20

Deleted��

Answered by Ar Brandon last updated on 06/Aug/20

$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}!\Rightarrow {\mathrm{lnA}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim ln}(\mathrm{n}!)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim ln}\underset{\mathrm{k}=0}{\prod ^{\mathrm{n}-1}}(\mathrm{n}-\mathrm{k})$$$${\mathrm{lnA}}_{\mathrm{n}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\ln (\mathrm{n}-\mathrm{k})=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}\cdot \frac{1}{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\ln \left(\mathrm{n}(1-\frac{\mathrm{k}}{\mathrm{n}})\right)$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}{\int }_0^1\ln \left(\mathrm{n}(1-\mathrm{x})\right)\mathrm{dx}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}{\int }_0^1\ln \left(\mathrm{tn}\right)\mathrm{dt}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left\{\mathrm{tn}(\ln \left(\mathrm{tn}\right)-1)\right\}}_0^1=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\{\mathrm{n}(\ln \left(\mathrm{n}\right)-1\}=+\mathrm{\infty }$$$${\mathrm{A}}_{\mathrm{n}}={\mathrm{e}}^{+\mathrm{\infty }}\ne \sqrt{2\pi }$$$$$$$$Ifeeltheremaybeaproblemwiththisquestion.$$

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