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

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

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

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 \dots \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}

\dots

\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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