Find-the-limit-of-n-4-n-as-n-approach-infinity-

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

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\:\:\:\:\frac{\mathrm{n}!}{\mathrm{4}^{\mathrm{n}} }\:\:\:\mathrm{as}\:\:\mathrm{n}\:\:\mathrm{approach}\:\mathrm{infinity} \\ $$

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 =+∞ .

$${let}\:{u}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\:\:\:{we}\:{have}\:{n}!\:\sim{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\left(\:{stirling}\:{formulae}\right)\:\Rightarrow \\ $$$$\frac{{n}!}{\mathrm{4}^{{n}} }\:\sim\:\left(\frac{{n}}{\mathrm{4}}\right)^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:=\:{e}^{{nln}\left(\frac{{n}}{\mathrm{4}}\right)−{n}} \sqrt{\mathrm{2}\pi{n}}\:\:={e}^{{n}\left\{{ln}\left(\frac{{n}}{\mathrm{4}}\right)−\mathrm{1}\right\}} \sqrt{\mathrm{2}\pi{n}}\:\rightarrow+\infty\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} {u}_{{n}} =+\infty\:. \\ $$

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 =+∞ .

$${another}\:{way}\:{let}\:{u}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\Rightarrow\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\:=\frac{\left({n}+\mathrm{1}\right)!}{\mathrm{4}^{{n}+\mathrm{1}} }\:.\frac{\mathrm{4}^{{n}} }{{n}!}\:=\frac{{n}+\mathrm{1}}{\mathrm{4}}\:\:\:\left({u}_{{n}} >\mathrm{0}\:\forall{n}\right)\:\Rightarrow \\ $$$${ln}\left({u}_{{n}+\mathrm{1}} \right)−{ln}\left({u}_{{n}} \right)\:={ln}\left({n}+\mathrm{1}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({ln}\left({u}_{{k}+\mathrm{1}} \right)−{ln}\left({u}_{{k}} \right)\right)={nln}\left({n}+\mathrm{1}\right)−\mathrm{2}{nln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${ln}\left({u}_{{n}} \right)\:={n}\left\{\:{ln}\left({n}+\mathrm{1}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)\right\}\:\rightarrow+\infty\:\Rightarrow{ln}\left({u}_{{n}} \right)\rightarrow+\infty\:\Rightarrow{lim}_{{n}\rightarrow\infty} {u}_{{n}} =+\infty\:. \\ $$

Commented by Tawa1 last updated on 23/Jun/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 23/Jun/19

$${you}\:{are}\:{most}\:{welcome}. \\ $$

Answered by MJS last updated on 23/Jun/19

((n!)/4^n )=((4!)/4^4 )×(5/4)×(6/4)×(7/4)×...×(n/4) ⇒ lim_(n→∞) ((n!)/4^n )=+∞

$$\frac{{n}!}{\mathrm{4}^{{n}} }=\frac{\mathrm{4}!}{\mathrm{4}^{\mathrm{4}} }×\frac{\mathrm{5}}{\mathrm{4}}×\frac{\mathrm{6}}{\mathrm{4}}×\frac{\mathrm{7}}{\mathrm{4}}×…×\frac{{n}}{\mathrm{4}} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}!}{\mathrm{4}^{{n}} }=+\infty \\ $$

Commented by Tawa1 last updated on 23/Jun/19

Sir, is the answer not zero ???. Just asking sir

$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{not}\:\mathrm{zero}\:???.\:\:\mathrm{Just}\:\mathrm{asking}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 23/Jun/19

Why do you start from n = 4 sir, and not n = 1

$$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{start}\:\mathrm{from}\:\:\mathrm{n}\:=\:\mathrm{4}\:\:\mathrm{sir},\:\mathrm{and}\:\mathrm{not}\:\:\mathrm{n}\:=\:\mathrm{1} \\ $$

Commented by MJS last updated on 23/Jun/19

((5!)/4^5 )=((15)/(128)) ((6!)/4^6 )=((45)/(256)) ((7!)/4^7 )=((315)/(1024)) ((8!)/4^8 )=((315)/(512)) ((9!)/4^9 )=((2835)/(2048)) ((10!)/4^(10) )=((14175)/(4096)) ... ((100!)/4^(100) )≈5.8×10^(97)

$$\frac{\mathrm{5}!}{\mathrm{4}^{\mathrm{5}} }=\frac{\mathrm{15}}{\mathrm{128}} \\ $$$$\frac{\mathrm{6}!}{\mathrm{4}^{\mathrm{6}} }=\frac{\mathrm{45}}{\mathrm{256}} \\ $$$$\frac{\mathrm{7}!}{\mathrm{4}^{\mathrm{7}} }=\frac{\mathrm{315}}{\mathrm{1024}} \\ $$$$\frac{\mathrm{8}!}{\mathrm{4}^{\mathrm{8}} }=\frac{\mathrm{315}}{\mathrm{512}} \\ $$$$\frac{\mathrm{9}!}{\mathrm{4}^{\mathrm{9}} }=\frac{\mathrm{2835}}{\mathrm{2048}} \\ $$$$\frac{\mathrm{10}!}{\mathrm{4}^{\mathrm{10}} }=\frac{\mathrm{14175}}{\mathrm{4096}} \\ $$$$… \\ $$$$\frac{\mathrm{100}!}{\mathrm{4}^{\mathrm{100}} }\approx\mathrm{5}.\mathrm{8}×\mathrm{10}^{\mathrm{97}} \\ $$

Commented by MJS last updated on 23/Jun/19

I showed that from (5/4) on we′re multiplicating by terms >1

$$\mathrm{I}\:\mathrm{showed}\:\mathrm{that}\:\mathrm{from}\:\frac{\mathrm{5}}{\mathrm{4}}\:\mathrm{on}\:\mathrm{we}'\mathrm{re}\:\mathrm{multiplicating} \\ $$$$\mathrm{by}\:\mathrm{terms}\:>\mathrm{1} \\ $$

Commented by Tawa1 last updated on 23/Jun/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Infinity}\:\mathrm{is}\:\mathrm{answer} \\ $$

Commented by MJS last updated on 23/Jun/19

you′re welcome maybe you had ((n!)/n^n ) in mind?

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$$$\mathrm{maybe}\:\mathrm{you}\:\mathrm{had}\:\frac{{n}!}{{n}^{{n}} }\:\mathrm{in}\:\mathrm{mind}? \\ $$

Commented by Tawa1 last updated on 23/Jun/19

I appreciate. now i know: lim_(∞→0) ((n!)/4^n ) = + ∞

$$\mathrm{I}\:\mathrm{appreciate}.\:\mathrm{now}\:\mathrm{i}\:\mathrm{know}:\:\:\:\:\:\:\underset{\infty\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{n}!}{\mathrm{4}^{\mathrm{n}} }\:\:=\:\:+\:\infty \\ $$

Commented by Tawa1 last updated on 23/Jun/19