Find the limit of ((n!)/4^n ) as n approach infinity


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Question Number 62626 by Tawa1 last updated on 23/Jun/19

$Find the limit of \frac{n!}{4^{n}} as n approach infinity$
Commented by mathmax by abdo last updated on 23/Jun/19
$let u_n =((n!)/4^n ) we have n! ∼n^n e^(−n) (√(2πn))( stirling formulae) ⇒ ((n!)/4^n ) ∼ ((n/4))^n e^(−n) (√(2πn)) = e^(nln((n/4))−n) (√(2πn)) =e^(n{ln((n/4))−1}) (√(2πn)) →+∞ (n→+∞) ⇒ lim_(n→∞) u_n =+∞ .$
$l e t u_{n} = \frac{n!}{4^{n}} w e h a v e n! \sim n^{n} e^{- n} \sqrt{2 π n} (s t i r l i n g f o r m u l a e) \Rightarrow$ $\frac{n!}{4^{n}} \sim {(\frac{n}{4})}^{n} e^{- n} \sqrt{2 π n} = e^{n l n (\frac{n}{4}) - n} \sqrt{2 π n} = e^{n {l n (\frac{n}{4}) - 1}} \sqrt{2 π n} \to + \infty (n \to + \infty) \Rightarrow$ ${l i m}_{n \to \infty} u_{n} = + \infty .$
Commented by mathmax by abdo last updated on 23/Jun/19
$another way let u_n =((n!)/4^n ) ⇒(u_(n+1) /u_n ) =(((n+1)!)/4^(n+1) ) .(4^n /(n!)) =((n+1)/4) (u_n >0 ∀n) ⇒ ln(u_(n+1) )−ln(u_n ) =ln(n+1)−2ln(2) ⇒ Σ_(k=0) ^(n−1) (ln(u_(k+1) )−ln(u_k ))=nln(n+1)−2nln(2) ⇒ ln(u_n ) =n{ ln(n+1)−2ln(2)} →+∞ ⇒ln(u_n )→+∞ ⇒lim_(n→∞) u_n =+∞ .$
$a n o t h e r w a y l e t u_{n} = \frac{n!}{4^{n}} \Rightarrow \frac{u_{n + 1}}{u_{n}} = \frac{(n + 1)!}{4^{n + 1}} . \frac{4^{n}}{n!} = \frac{n + 1}{4} (u_{n} > 0 \forall n) \Rightarrow$ $l n (u_{n + 1}) - l n (u_{n}) = l n (n + 1) - 2 l n (2) \Rightarrow$ $\sum_{k = 0}^{n - 1} (l n (u_{k + 1}) - l n (u_{k})) = n l n (n + 1) - 2 n l n (2) \Rightarrow$ $l n (u_{n}) = n {l n (n + 1) - 2 l n (2)} \to + \infty \Rightarrow l n (u_{n}) \to + \infty \Rightarrow {l i m}_{n \to \infty} u_{n} = + \infty .$
Commented by Tawa1 last updated on 23/Jun/19

$God bless you sir$
Commented by mathmax by abdo last updated on 23/Jun/19

$y o u a r e m o s t w e l c o m e .$
	Answered by MJS last updated on 23/Jun/19

	$\frac{n!}{4^{n}} = \frac{4!}{4^{4}} \times \frac{5}{4} \times \frac{6}{4} \times \frac{7}{4} \times . . . \times \frac{n}{4}$ $\Rightarrow \lim_{n \to \infty} \frac{n!}{4^{n}} = + \infty$
	Commented by Tawa1 last updated on 23/Jun/19

	$Sir, is the answer not zero ? ? ? . Just asking sir$
	Commented by Tawa1 last updated on 23/Jun/19

	$Why do you start from n = 4 sir, and not n = 1$
	Commented by MJS last updated on 23/Jun/19

	$\frac{5!}{4^{5}} = \frac{15}{128}$ $\frac{6!}{4^{6}} = \frac{45}{256}$ $\frac{7!}{4^{7}} = \frac{315}{1024}$ $\frac{8!}{4^{8}} = \frac{315}{512}$ $\frac{9!}{4^{9}} = \frac{2835}{2048}$ $\frac{10!}{4^{10}} = \frac{14175}{4096}$ $. . .$ $\frac{100!}{4^{100}} \approx 5.8 \times 10^{97}$
	Commented by MJS last updated on 23/Jun/19

	$I showed that from \frac{5}{4} on {we}^{'} re multiplicating$ $by terms > 1$
	Commented by Tawa1 last updated on 23/Jun/19

	$God bless you sir . Infinity is answer$
	Commented by MJS last updated on 23/Jun/19

	${you}^{'} re welcome$ $maybe you had \frac{n!}{n^{n}} in mind ?$
	Commented by Tawa1 last updated on 23/Jun/19

	$I appreciate . now i know : \lim_{\infty \to 0} \frac{n!}{4^{n}} = + \infty$
	Commented by Tawa1 last updated on 23/Jun/19

	$The answer is this is zero sir ?$
	Commented by Tawa1 last updated on 23/Jun/19

	$\lim_{n \to \infty} \frac{n!}{n^{n}} = 0$
	Commented by MJS last updated on 23/Jun/19

	$y e s$
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