Question-106591

Facebook Tweet Pin

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

Commented by Her_Majesty last updated on 06/Aug/20

$$\infty\:{is}\:{not}\:{a}\:{number} \\ $$$$\underset{{x}\rightarrow\infty} {{lim}x}!\:\neq\sqrt{\mathrm{2}\pi} \\ $$

Commented by mathdave last updated on 06/Aug/20

$${yes}\:{but}\:{can}\:{b}\:{prove} \\ $$

Commented by Her_Majesty last updated on 06/Aug/20

it′s simply wrong. x!≈((x/e))^x (√(2πx)) for x→+∞ we get x!→+∞

$${it}'{s}\:{simply}\:{wrong}. \\ $$$${x}!\approx\left(\frac{{x}}{\mathrm{e}}\right)^{{x}} \sqrt{\mathrm{2}\pi{x}}\:{for}\:{x}\rightarrow+\infty\:{we}\:{get}\:{x}!\rightarrow+\infty \\ $$

Commented by Her_Majesty last updated on 07/Aug/20

$${I}'{m}\:{not}\:{a}\:{Sir}\:\mathrm{beware}!!!\:{I}'{m}\:{a}\:{Lady} \\ $$

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

$${what}\:{are}\:{you}\:{talking}\:{about}? \\ $$

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

Deleted��

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

A_n =lim_(n→∞) n!⇒lnA_n =lim_(n→∞) ln(n!)=lim_(n→∞) lnΠ_(k=0) ^(n−1) (n−k) lnA_n =lim_(n→∞) Σ_(k=0) ^(n−1) ln(n−k)=lim_(n→∞) n∙(1/n)Σ_(k=0) ^(n−1) ln(n(1−(k/n))) =lim_(n→∞) n∫_0 ^1 ln(n(1−x))dx=lim_(n→∞) n∫_0 ^1 ln(tn)dt =lim_(n→∞) {tn(ln(tn)−1)}_0 ^1 =lim_(n→∞) {n(ln(n)−1}=+∞ A_n =e^(+∞) ≠(√(2π)) I feel there maybe a problem with this question.

$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}!\Rightarrow\mathrm{lnA}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}ln}\left(\mathrm{n}!\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}ln}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{n}−\mathrm{k}\right) \\ $$$$\mathrm{lnA}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{ln}\left(\mathrm{n}−\mathrm{k}\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\centerdot\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{ln}\left(\mathrm{n}\left(\mathrm{1}−\frac{\mathrm{k}}{\mathrm{n}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{n}\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{dx}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{tn}\right)\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{tn}\left(\mathrm{ln}\left(\mathrm{tn}\right)−\mathrm{1}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{n}\left(\mathrm{ln}\left(\mathrm{n}\right)−\mathrm{1}\right\}=+\infty\right. \\ $$$$\mathrm{A}_{\mathrm{n}} =\mathrm{e}^{+\infty} \neq\sqrt{\mathrm{2}\pi} \\ $$$$ \\ $$$${I}\:{feel}\:{there}\:{maybe}\:{a}\:{problem}\:{with}\:{this}\:{question}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply