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Reposting-question-181462-lim-n-2-3-1-3-4-1-4-n-1-n-1-n-2n-1-1-n-1-1-2e-

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Question Number 181553 by Frix last updated on 26/Nov/22
Reposting question 181462 lim_(n→∞) ((((√(2!))×((3!))^(1/3) ×((4!))^(1/4) ×...×((n!))^(1/n) ))^(1/n) /( (((2n+1)!!))^(1/(n+1)) ))=^? (1/(2e))
$$\mathrm{Reposting}\:\mathrm{question}\:\mathrm{181462} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}}{\:\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2e}} \\ $$
Answered by aleks041103 last updated on 26/Nov/22
b_n =((((√(2!))×((3!))^(1/3) ×((4!))^(1/4) ×...×((n!))^(1/n) ))^(1/n) /( (((2n+1)!!))^(1/(n+1)) ))=((((√(2!))×((3!))^(1/3) ×((4!))^(1/4) ×...×((n!))^(1/n) )/(((2n+1)!!)^(n/(n+1)) )))^(1/n) =(a_n )^(1/n) a_n =(((√(2!))×((3!))^(1/3) ×((4!))^(1/4) ×...×((n!))^(1/n) )/(((2n+1)!!)^(n/(n+1)) )) lim_(n→∞) (a_n )^(1/n) =lim_(n→∞) (a_(n+1) /a_n ) (a_(n+1) /a_n )=(((n+1)!))^(1/(n+1)) ((((2n+1)!!)^(n/(n+1)) )/(((2n+3)!!)^((n+1)/(n+2)) )) (2n+1)!!=(((2n+1)!)/((2n)!!))=(((2n+1)!)/(2^n n!)) stirling approx x!∼x^x e^(−x) (√(2πx))=(√(2π)) x^(x+1/2) e^(−x) ((2n+1)!!)^(n/(n+1)) =((((2n+1)!)/(2^n n!)))^(n/(n+1)) ∼((((2n+1)^(2n+3/2) e^(−2n−1) )/(2^n n^(n+1/2) e^(−n) )))^(n/(n+1)) = =n^n (1+(1/(2n)))^((n(2n+3/2))/((n+1))) e^(−n) 2^(((n(2n+3/2))/(n+1))−(n^2 /(n+1))) ⇒((2n+1)!!)^(n/(n+1)) ∼e^(−n) n^n 2^((n(n+3/2))/(n+1)) (1+(1/(2n)))^(2n((n+3/4)/(n+1))) ∼e^(1−n) (2n)^n ⇒(a_(n+1) /a_n )=(((n+1)!))^(1/(n+1)) ((((2n+1)!!)^(n/(n+1)) )/(((2n+3)!!)^((n+1)/(n+2)) )) ∼((e^(1−n) 2^n n^n )/(e^(−n) 2^(n+1) (n+1)^(n+1) ))((n+1)^(n+1) e^(−n−1) (√n))^(1/(n+1)) = =(e/(2(n+1)(1+(1/n))^n ))(n+1)e^(−1) =(1/(2e)) ⇒Ans. (1/(2e))
$${b}_{{n}} =\frac{\sqrt[{{n}}]{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}}{\:\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}=\sqrt[{{n}}]{\frac{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}{\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} }}=\sqrt[{{n}}]{{a}_{{n}} } \\ $$$${a}_{{n}} =\frac{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}{\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} } \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\sqrt[{{n}}]{{a}_{{n}} }=\underset{{n}\rightarrow\infty} {{lim}}\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} } \\ $$$$ \\ $$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!}\frac{\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} }{\left(\left(\mathrm{2}{n}+\mathrm{3}\right)!!\right)^{\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}} } \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\left(\mathrm{2}{n}\right)!!}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!} \\ $$$${stirling}\:{approx} \\ $$$${x}!\sim{x}^{{x}} {e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}=\sqrt{\mathrm{2}\pi}\:{x}^{{x}+\mathrm{1}/\mathrm{2}} {e}^{−{x}} \\ $$$$\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} =\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}\right)^{\frac{{n}}{{n}+\mathrm{1}}} \sim\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{3}/\mathrm{2}} {e}^{−\mathrm{2}{n}−\mathrm{1}} }{\mathrm{2}^{{n}} {n}^{{n}+\mathrm{1}/\mathrm{2}} {e}^{−{n}} }\right)^{\frac{{n}}{{n}+\mathrm{1}}} = \\ $$$$={n}^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)^{\frac{{n}\left(\mathrm{2}{n}+\mathrm{3}/\mathrm{2}\right)}{\left({n}+\mathrm{1}\right)}} {e}^{−{n}} \mathrm{2}^{\frac{{n}\left(\mathrm{2}{n}+\mathrm{3}/\mathrm{2}\right)}{{n}+\mathrm{1}}−\frac{{n}^{\mathrm{2}} }{{n}+\mathrm{1}}} \\ $$$$\Rightarrow\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} \sim{e}^{−{n}} {n}^{{n}} \mathrm{2}^{\frac{{n}\left({n}+\mathrm{3}/\mathrm{2}\right)}{{n}+\mathrm{1}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)^{\mathrm{2}{n}\frac{{n}+\mathrm{3}/\mathrm{4}}{{n}+\mathrm{1}}} \sim{e}^{\mathrm{1}−{n}} \left(\mathrm{2}{n}\right)^{{n}} \\ $$$$\Rightarrow\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!}\frac{\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}} }{\left(\left(\mathrm{2}{n}+\mathrm{3}\right)!!\right)^{\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}} } \\ $$$$\sim\frac{{e}^{\mathrm{1}−{n}} \mathrm{2}^{{n}} {n}^{{n}} }{{e}^{−{n}} \mathrm{2}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\left(\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{−{n}−\mathrm{1}} \sqrt{{n}}\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} = \\ $$$$=\frac{{e}}{\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} }\left({n}+\mathrm{1}\right){e}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{e}} \\ $$$$\Rightarrow{Ans}.\:\frac{\mathrm{1}}{\mathrm{2}{e}} \\ $$$$ \\ $$
Commented by Frix last updated on 26/Nov/22
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Answered by Frix last updated on 26/Nov/22
My idea: (1) ((n!))^(1/n) =((Π_(k=1) ^n k))^(1/n) <((Σ_(k=1) ^n k)/n)=((n+1)/2) ⇒ ((n+1)/(2((n!))^(1/n) ))=c_n >1 lim_(n→∞) c_n =lim_(n→∞) ((n+1)/(2(n/e)((2πn))^(1/(2n)) )) =(e/2) [easy to see] ⇒ ((n!))^(1/n) ∼((n+1)/e) for high values of n ⇒ ((Π_(k=1) ^n ((k!))^(1/k) ))^(1/n) ∼((Π_(k=1) ^n ((k+1)/e)))^(1/n) =((((n+1)n!)/e^n ))^(1/n) = =((((n+1))^(1/n) ((n!))^(1/n) )/e)=((((n+1))^(1/n) (n+1))/e^2 ) (2) (((2n+1)!!))^(1/(n+1)) ∼(((2n−1)!!))^(1/n) =((((2n)!)/(2^n n!)))^(1/n) = =((((2n)!))^(1/n) /(2((n!))^(1/n) ))=((e (((2n!)))^(1/n) )/(2(n+1)))=((2n^2 ((4πn))^(1/(2n)) )/(e(n+1))) Now we have lim_(n→∞) ((((√(2!))×((3!))^(1/3) ×((4!))^(1/4) ×...×((n!))^(1/n) ))^(1/n) /( (((2n+1)!!))^(1/(n+1)) ))= =lim_(n→∞) ((((n+1))^(1/n) (n+1)^2 )/(2en^2 ((4πn))^(1/(2n)) )) =(1/(2e)) [easy to see] Is this correct?
$$\mathrm{My}\:\mathrm{idea}: \\ $$$$\left(\mathrm{1}\right) \\ $$$$\sqrt[{{n}}]{{n}!}=\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{k}}<\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}}{{n}}=\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{{n}+\mathrm{1}}{\mathrm{2}\sqrt[{{n}}]{{n}!}}={c}_{{n}} >\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{c}_{{n}} \:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}+\mathrm{1}}{\mathrm{2}\frac{{n}}{\mathrm{e}}\sqrt[{\mathrm{2}{n}}]{\mathrm{2}\pi{n}}}\:=\frac{\mathrm{e}}{\mathrm{2}}\:\left[\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\right] \\ $$$$\Rightarrow\:\sqrt[{{n}}]{{n}!}\sim\frac{{n}+\mathrm{1}}{\mathrm{e}}\:\mathrm{for}\:\mathrm{high}\:\mathrm{values}\:\mathrm{of}\:{n} \\ $$$$\Rightarrow\:\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\sqrt[{{k}}]{{k}!}}\sim\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{k}+\mathrm{1}}{\mathrm{e}}}=\sqrt[{{n}}]{\frac{\left({n}+\mathrm{1}\right){n}!}{\mathrm{e}^{{n}} }}= \\ $$$$=\frac{\sqrt[{{n}}]{{n}+\mathrm{1}}\sqrt[{{n}}]{{n}!}}{\mathrm{e}}=\frac{\sqrt[{{n}}]{{n}+\mathrm{1}}\left({n}+\mathrm{1}\right)}{\mathrm{e}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right) \\ $$$$\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}\sim\sqrt[{{n}}]{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}=\sqrt[{{n}}]{\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} {n}!}}= \\ $$$$=\frac{\sqrt[{{n}}]{\left(\mathrm{2}{n}\right)!}}{\mathrm{2}\sqrt[{{n}}]{{n}!}}=\frac{\mathrm{e}\:\sqrt[{{n}}]{\left(\mathrm{2}{n}!\right)}}{\mathrm{2}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{2}{n}^{\mathrm{2}} \:\sqrt[{\mathrm{2}{n}}]{\mathrm{4}\pi{n}}}{\mathrm{e}\left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}}{\:\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}= \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{{n}+\mathrm{1}}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2e}{n}^{\mathrm{2}} \:\sqrt[{\mathrm{2}{n}}]{\mathrm{4}\pi{n}}}\:=\frac{\mathrm{1}}{\mathrm{2e}}\:\left[\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\right] \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}? \\ $$

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