calculate lim_(n→+∞) (((n!)^n )/n^(n!) )


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Question Number 62659 by mathmax by abdo last updated on 24/Jun/19

$c a l c u l a t e {l i m}_{n \to + \infty} \frac{{(n!)}^{n}}{n^{n!}}$
Commented by abdo mathsup 649 cc last updated on 24/Jun/19
$let A_n =(((n!)^n )/n^(n!) ) ⇒ln(A_n ) =nln(n!)−n!ln(n) we have for n∈V(+∞) n! ∼n^(n ) e^(−n) (√(2πn)) ⇒ ln(n!)∼nln(n)+n +(1/2)ln(2πn) ⇒ ln(A_n )∼ n^2 ln(n)+n^(2 ) +(n/2)ln(2πn)−n^(n ) e^(−n) (√(2πn))ln(n) =n^n { ((ln(n))/n^(n−2) ) +(1/n^(n−2) ) + ((ln(2πn))/(2n^(n−1) )) −e^(−n) (((√(2πn))ln(n))/n^n )}→0 (n→+∞) ⇒lim_(n→+∞) A_n =1 .$
$l e t A_{n} = \frac{{(n!)}^{n}}{n^{n!}} \Rightarrow l n (A_{n}) = n l n (n!) - n! l n (n)$ $w e h a v e f o r n \in V (+ \infty) n! \sim n^{n} e^{- n} \sqrt{2 π n} \Rightarrow$ $l n (n!) \sim n l n (n) + n + \frac{1}{2} l n (2 π n) \Rightarrow$ $l n (A_{n}) \sim n^{2} l n (n) + n^{2} + \frac{n}{2} l n (2 π n) - n^{n} e^{- n} \sqrt{2 π n} l n (n)$ $= n^{n} {\frac{l n (n)}{n^{n - 2}} + \frac{1}{n^{n - 2}} + \frac{l n (2 π n)}{2 n^{n - 1}} - e^{- n} \frac{\sqrt{2 π n} l n (n)}{n^{n}}} \to 0$ $(n \to + \infty) \Rightarrow {l i m}_{n \to + \infty} A_{n} = 1 .$
Commented by abdo mathsup 649 cc last updated on 24/Jun/19

$l n (n!) \sim n l n (n) - n + \frac{1}{2} l n (2 π n)$
Commented by MJS last updated on 24/Jun/19

$n \ln (n^{n} e^{- n} \sqrt{2 π n}) = (n^{2} + \frac{n}{2}) \ln n - n^{2} + \frac{\ln 2 π}{2} n$ $(n^{2} + \frac{n}{2}) \ln n - n^{2} + \frac{\ln 2 π}{2} n - n^{n} e^{- n} \sqrt{2 π n} \ln n =$ $= (n^{2} + \frac{n}{2} - {(\frac{n}{e})}^{n} \sqrt{2 π n}) \ln n - n^{2} + \frac{\ln 2 π}{2} n$ $and because of - {(\frac{n}{e})}^{n} \sqrt{2 π n} \ln n this approaches$ $not 0 but - \infty \Rightarrow \lim_{n \to + \infty} A_{n} = e^{- \infty} = 0$
	Answered by MJS last updated on 24/Jun/19

	$\frac{{(n!)}^{n}}{n^{n!}} = f (n)$ $f (1) = \frac{1}{1} = 1$ $f (2) = \frac{4}{4} = 1$ $f (3) = \frac{216}{729} = \frac{8}{27}$ $f (4) = \frac{331776}{281474976710656} = \frac{81}{68719476736}$ $. . .$ $f (\infty) = 0$
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