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fnd-lim-n-2n-n-n-n-1-n-

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Question Number 107289 by mathmax by abdo last updated on 09/Aug/20
fnd lim_(n→+∞) ((((2n)!)/(n^n n!)))^(1/n)
fndlimn→+∞((2n)!nnn!)1n
Answered by Ar Brandon last updated on 13/Aug/20
A_n = ((((2n)!)/(n^n n!)))^(1/n) lnA_n =ln ((((2n)!)/(n^n n!)))^(1/n) =(1/n)[ln(2n)!−ln(n^n )−ln(n!)] =(1/n)[lnΠ_(k=0) ^(2n−1) (2n−k)−lnΠ_(k=0) ^(n−1) (n−k)]−ln(n) =(1/n)[Σ_(k=0) ^(2n−1) ln(2n−k)−Σ_(k=0) ^(n−1) ln(n−k)]−ln(n) =(1/n)[2n∙ln(n)+Σ_(k=0) ^(2n−1) ln(2−(k/n))−n∙ln(n)−Σ_(k=0) ^(n−1) ln(1−(k/n))]−ln(n) lim_(n→∞) lnA_n =∫_0 ^2 ln(2−x)dx−∫_0 ^1 ln(1−x)dx =∫_0 ^2 ln(t_1 )dt_1 −∫_0 ^1 ln(t_2 )dt_2 =[tlnt−t]_1 ^2 =2ln2−1 lim_(n→∞) A_n =e^(2ln2−1) =(4/e)
An=((2n)!nnn!)1nlnAn=ln((2n)!nnn!)1n=1n[ln(2n)!−ln(nn)−ln(n!)]=1n[ln∏2n−1k=0(2n−k)−ln∏n−1k=0(n−k)]−ln(n)=1n[∑2n−1k=0ln(2n−k)−∑n−1k=0ln(n−k)]−ln(n)=1n[2n⋅ln(n)+∑2n−1k=0ln(2−kn)−n⋅ln(n)−∑n−1k=0ln(1−kn)]−ln(n)Double subscripts: use braces to clarify=∫02ln(t1)dt1−∫01ln(t2)dt2=[tlnt−t]12=2ln2−1Double subscripts: use braces to clarify

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