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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))

Repostingquestion181462limn→∞2!×3!3×4!4×...×n!nn(2n+1)!!n+1=?12e

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))

bn=2!×3!3×4!4×...×n!nn(2n+1)!!n+1=2!×3!3×4!4×...×n!n((2n+1)!!)nn+1n=annan=2!×3!3×4!4×...×n!n((2n+1)!!)nn+1limn→∞ann=limn→∞an+1anan+1an=(n+1)!n+1((2n+1)!!)nn+1((2n+3)!!)n+1n+2(2n+1)!!=(2n+1)!(2n)!!=(2n+1)!2nn!stirlingapproxx!∼xxe−x2πx=2πxx+1/2e−x((2n+1)!!)nn+1=((2n+1)!2nn!)nn+1∼((2n+1)2n+3/2e−2n−12nnn+1/2e−n)nn+1==nn(1+12n)n(2n+3/2)(n+1)e−n2n(2n+3/2)n+1−n2n+1⇒((2n+1)!!)nn+1∼e−nnn2n(n+3/2)n+1(1+12n)2nn+3/4n+1∼e1−n(2n)n⇒an+1an=(n+1)!n+1((2n+1)!!)nn+1((2n+3)!!)n+1n+2∼e1−n2nnne−n2n+1(n+1)n+1((n+1)n+1e−n−1n)1n+1==e2(n+1)(1+1n)n(n+1)e−1=12e⇒Ans.12e

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?

Myidea:(1)n!n=∏nk=1kn<∑nk=1kn=n+12⇒n+12n!n=cn>1limn→∞cn=limn→∞n+12ne2πn2n=e2[easytosee]⇒n!n∼n+1eforhighvaluesofn⇒∏nk=1k!kn∼∏nk=1k+1en=(n+1)n!enn==n+1nn!ne=n+1n(n+1)e2(2)(2n+1)!!n+1∼(2n−1)!!n=(2n)!2nn!n==(2n)!n2n!n=e(2n!)n2(n+1)=2n24πn2ne(n+1)Nowwehavelimn→∞2!×3!3×4!4×...×n!nn(2n+1)!!n+1==limn→∞n+1n(n+1)22en24πn2n=12e[easytosee]Isthiscorrect?

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