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Question Number 60981 by necx1 last updated on 28/May/19

Commented by Prithwish sen last updated on 28/May/19

Let A=lim_(x→∞) (((x!)/x^x ))^(1/x) ∴lnA=lim_(x→∞) (1/x)ln(((1.2.3...........x)/(x.x.x............x))) =lim_(x→∞) (1/x)(ln(1/x) +ln(2/x) + ........ln(x/x) ) =lim_(x→∞) (1/x)Σ_(r=1) ^x ln(r/x) =∫_0 ^1 lnydy=(ylny−y)_0 ^1 =−1 ∴ A=e^(−1) =(1/e)

$$\mathrm{Let}\: \\ $$$$\mathrm{A}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \left(\frac{\mathrm{x}!}{\mathrm{x}^{\mathrm{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\therefore\mathrm{lnA}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}...........\mathrm{x}}{\mathrm{x}.\mathrm{x}.\mathrm{x}............\mathrm{x}}\right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{x}}\left(\mathrm{ln}\frac{\mathrm{1}}{\mathrm{x}}\:+\mathrm{ln}\frac{\mathrm{2}}{\mathrm{x}}\:+\:........\mathrm{ln}\frac{\mathrm{x}}{\mathrm{x}}\:\right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{x}}\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{x}} {\sum}}\mathrm{ln}\frac{\mathrm{r}}{\mathrm{x}} \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{lnydy}=\left(\mathrm{ylny}−\mathrm{y}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\right)}} \\ $$$$=−\mathrm{1} \\ $$$$\therefore\:\mathrm{A}=\mathrm{e}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$

Commented by tanmay last updated on 28/May/19

bah darun excellent...

$${bah}\:{darun}\:{excellent}... \\ $$

Commented by Prithwish sen last updated on 28/May/19

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 28/May/19

Extremely great Sir, thanks a lot.

$${Extremely}\:{great}\:{Sir},\:{thanks}\:{a}\:{lot}. \\ $$

Commented by Prithwish sen last updated on 28/May/19

welcome.

$$\mathrm{welcome}.\: \\ $$

Commented by Mr X pcx last updated on 28/May/19

stirling gormulae n! ∼ n^n e^(−n) (√(2πn)) or x! ∼x^x e^(−x) (√(2πx)) ⇒((x!)/x^x ) ∼ e^(−x) (√(2πx)) ⇒ (((x!)/x^x ))^(1/x) =(e^(−x) (√(2πx)))^(1/x) =A(x) ln(A(x)) =(1/x)ln(e^(−x) (√(2πx))) =−1 +(1/(2x)){ln(2π) +ln(x)} =−1 +((ln(2π))/(2x)) +((ln(x))/(2x)) →−1(x→+∞) ⇒lim_(x→+∞) A(x)=e^(−1) =(1/e) so lim_(x→+∞) (((x!)/x^x ))^(1/x) =(1/e) .

$${stirling}\:{gormulae} \\ $$$${n}!\:\sim\:{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\:\:\:\:{or}\: \\ $$$${x}!\:\sim{x}^{{x}} \:{e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\:\Rightarrow\frac{{x}!}{{x}^{{x}} }\:\sim\:{e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\:\Rightarrow \\ $$$$\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \:\:=\left({e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\right)^{\frac{\mathrm{1}}{{x}}} \:={A}\left({x}\right) \\ $$$${ln}\left({A}\left({x}\right)\right)\:=\frac{\mathrm{1}}{{x}}{ln}\left({e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\right) \\ $$$$=−\mathrm{1}\:\:+\frac{\mathrm{1}}{\mathrm{2}{x}}\left\{{ln}\left(\mathrm{2}\pi\right)\:+{ln}\left({x}\right)\right\} \\ $$$$=−\mathrm{1}\:+\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}{x}}\:+\frac{{ln}\left({x}\right)}{\mathrm{2}{x}}\:\rightarrow−\mathrm{1}\left({x}\rightarrow+\infty\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)={e}^{−\mathrm{1}} \:=\frac{\mathrm{1}}{{e}} \\ $$$${so}\:{lim}_{{x}\rightarrow+\infty} \:\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \:=\frac{\mathrm{1}}{{e}}\:. \\ $$

Answered by MJS last updated on 28/May/19

tried with Stirling′s formula x!≈((x/e))^x (√(2πx)) (((x!)/x^x ))^(1/x) =((x!^(1/x) )/x)≈(1/x)(((x/e))^x (√(2πx)))^(1/x) =(1/x)×(x/e)(2πx)^(1/(2x)) = =(1/e)×2^(1/(2x)) ×π^(1/(2x)) ×x^(1/(2x)) lim_(x→∞) (1/e)×2^(1/(2x)) ×π^(1/(2x)) ×x^(1/(2x)) =(1/e) this seems right, computing shows (((100!)/(100^(100) )))^(1/(100)) ≈.379927 (((200!)/(200^(200) )))^(1/(200)) ≈.374502 (((300!)/(300^(300) )))^(1/(300)) ≈.372533 (((400!)/(400^(400) )))^(1/(400)) ≈.371498 (1/e)≈.367879

$$\mathrm{tried}\:\mathrm{with}\:\mathrm{Stirling}'\mathrm{s}\:\mathrm{formula} \\ $$$${x}!\approx\left(\frac{{x}}{\mathrm{e}}\right)^{{x}} \sqrt{\mathrm{2}\pi{x}} \\ $$$$\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} =\frac{{x}!^{\frac{\mathrm{1}}{{x}}} }{{x}}\approx\frac{\mathrm{1}}{{x}}\left(\left(\frac{{x}}{\mathrm{e}}\right)^{{x}} \sqrt{\mathrm{2}\pi{x}}\right)^{\frac{\mathrm{1}}{{x}}} =\frac{\mathrm{1}}{{x}}×\frac{{x}}{\mathrm{e}}\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} ×\pi^{\frac{\mathrm{1}}{\mathrm{2}{x}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{e}}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} ×\pi^{\frac{\mathrm{1}}{\mathrm{2}{x}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$$$\mathrm{this}\:\mathrm{seems}\:\mathrm{right},\:\mathrm{computing}\:\mathrm{shows} \\ $$$$\left(\frac{\mathrm{100}!}{\mathrm{100}^{\mathrm{100}} }\right)^{\frac{\mathrm{1}}{\mathrm{100}}} \approx.\mathrm{379927} \\ $$$$\left(\frac{\mathrm{200}!}{\mathrm{200}^{\mathrm{200}} }\right)^{\frac{\mathrm{1}}{\mathrm{200}}} \approx.\mathrm{374502} \\ $$$$\left(\frac{\mathrm{300}!}{\mathrm{300}^{\mathrm{300}} }\right)^{\frac{\mathrm{1}}{\mathrm{300}}} \approx.\mathrm{372533} \\ $$$$\left(\frac{\mathrm{400}!}{\mathrm{400}^{\mathrm{400}} }\right)^{\frac{\mathrm{1}}{\mathrm{400}}} \approx.\mathrm{371498} \\ $$$$\frac{\mathrm{1}}{\mathrm{e}}\approx.\mathrm{367879} \\ $$

Commented by tanmay last updated on 28/May/19

bah darun excellent...one problem but solution three method...

$${bah}\:{darun}\:{excellent}...{one}\:{problem}\:{but}\:{solution} \\ $$$${three}\:{method}... \\ $$

Answered by tanmay last updated on 28/May/19

lim_(x→∞) (a_(x+1) /a_x ) =lim_(x→∞) (({(((x+1)!)/((x+1)^(x+1) ))})/((((x!)/x^x )))) =lim_(x→∞) (((x+1)!)/(x!))×(1/(((x+1)^(x+1) )/x^x )) =lim_(x→∞) (x+1)×(1/((x+1)(((x+1)/x))^x )) =lim_(x→∞) (1/((1+(1/x))^x )) now t=(1/x) y=lim_(t→0) (1/((1+t)^(1/t) )) lny=lim_(t→0) [((−ln(1+t))/t)] lny=−1 y=e^(−1) so lim_(x→∞) (((x!)/x^x ))^(1/x) =e^(−1) →it is the answer using cauchy second theorem on limit lim_(n→∞) (a_n )^(1/n) =l if lim_(n→∞) (a_(n+1) /a_n )=l

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{x}+\mathrm{1}} }{{a}_{{x}} } \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\left\{\frac{\left({x}+\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} }\right\}}{\left(\frac{{x}!}{{x}^{{x}} }\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{1}\right)!}{{x}!}×\frac{\mathrm{1}}{\frac{\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} }{{x}^{{x}} }} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({x}+\mathrm{1}\right)×\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}} } \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} } \\ $$$${now}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$${y}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{{t}}} } \\ $$$${lny}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{−{ln}\left(\mathrm{1}+{t}\right)}{{t}}\right] \\ $$$${lny}=−\mathrm{1} \\ $$$${y}={e}^{−\mathrm{1}} \\ $$$${so}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{−\mathrm{1}} \rightarrow{it}\:{is}\:{the}\:{answer} \\ $$$${using}\:{cauchy}\:{second}\:{theorem}\:{on}\:{limit} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({a}_{{n}} \right)^{\frac{\mathrm{1}}{{n}}} ={l} \\ $$$${if}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }={l} \\ $$

Commented by necx1 last updated on 28/May/19

God bless you sir

$${God}\:{bless}\:{you}\:{sir} \\ $$

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