l_n i_ m_∞ ((((n!))/((nm)^n )))^(1/n)


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 34092 by rahul 19 last updated on 30/Apr/18

$\underset{n}{l} \underset{}{i} \underset{\infty}{m} {(\frac{(n!)}{{(n m)}^{n}})}^{\frac{1}{n}}$
Commented by MJS last updated on 30/Apr/18

$Stirling approximation :$ $n! \approx {(\frac{n}{e})}^{n} \sqrt{2 π n}$ $\lim_{x \to \infty} {(\frac{n}{e})}^{n} \sqrt{2 π n} = n!$ $. . . the rest should be easy . . .$
Commented by rahul 19 last updated on 30/Apr/18

$p l s d o t h i s b y^{'} s t i r l i n g {a p p r o x i m a t i o n}^{'}$ $a n d a l s o d o e x p l a i n {w h a t}^{'} s t h a t .$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

$t h e r e i s a m e t h o d t o s o l v e t h i s t y p e o f p r o b l e m$ $l e t m e s h a r e t h e o r e m f i r s t a l o n g w i t h s o m e$ $p r o b l e m t h e n i s h a l l s o l v e i t$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18
$x_n =((n!)/((nm)^n )) so x_(n+1) =(((n+1)!)/({(n+1)m}^(n+1) )) (x_n /(x_(n+1) ))=((n!)/((n+1)!)).(({(n+1)m}^(n+1) )/((nm)^n )) =(1/(n+1)).(({{(n+1)m}^n )/((nm)^(n ) )).{(n+1)m} =m.(((n+1)^n )/n^n ) (x_(n+1) /x_n )=(1/m).(n^n /((n+1)^n )) =(1/m).(1/((1+((1 )/n))^n )) lim n→∞(x_(n+1) /x_n ) =(1/(e.m)) that is required ans of the question$
$x_{n} = \frac{n!}{{(n m)}^{n}} s o x_{n + 1} = \frac{(n + 1)!}{{(n + 1) m}^{n + 1}}$ $\frac{x_{n}}{x_{n + 1}} = \frac{n!}{(n + 1)!} . \frac{{(n + 1) m}^{n + 1}}{{(n m)}^{n}}$ $= \frac{1}{n + 1} . \frac{{{(n + 1) m}^{n}}{{(n m)}^{n}} . {(n + 1) m}$ $= m . \frac{{(n + 1)}^{n}}{n^{n}}$ $\frac{x_{n + 1}}{x_{n}} = \frac{1}{m} . \frac{n^{n}}{{(n + 1)}^{n}}$ $= \frac{1}{m} . \frac{1}{{(1 + \frac{1}{n})}^{n}}$ $l i m n \to \infty \frac{x_{n + 1}}{x_{n}}$ $= \frac{1}{e . m}$ $t h a t i s r e q u i r e d a n s o f t h e q u e s t i o n$
Commented by rahul 19 last updated on 01/May/18

$T h a n k y o u s i r .$ $w e l l, i t h i n k s t i r l i n g a p r r o x i m a t i o n$ $i s b e t t e r (a s i t i s s h o r t) . A n y w a y s,$ ${i t}^{'} s a l s o a n i c e m e t h o d .$
Commented by rahul 19 last updated on 01/May/18
pls do post the solutions of the given 5 Ques. also by this theorem.
Commented by math khazana by abdo last updated on 02/May/18

$l e t p u t u_{n} = {(\frac{(n + 1) (n + 2) . . . (n + n)}{n})}^{\frac{1}{n}}$ $u_{n} = {(\frac{n^{n} (1 + \frac{1}{n}) (1 + \frac{2}{n}) . . . . (1 + \frac{n}{n})}{n})}^{\frac{1}{n}}$ $= n^{\frac{n - 1}{n}} {(\prod_{k = 1}^{n} (1 + \frac{k}{n}))}^{\frac{1}{n}} = n^{1 - \frac{1}{n}} {(\prod_{k = 1}^{n} (1 + \frac{k}{n}))}^{\frac{1}{n}}$ $\Rightarrow l n (u_{n}) = (1 - \frac{1}{n}) l n (n) + \frac{1}{n} \sum_{k = 1}^{n} l n (1 + \frac{k}{n})$ $b u t {l i m}_{n \to + \infty} \frac{1}{n} \sum_{k = 1}^{n} l n (1 + \frac{k}{n}) = \int_{0}^{1} l n (1 + x) d x$ ${l i m}_{n \to + \infty} (1 - \frac{1}{n}) l n (n) = + \infty \Rightarrow {l i m}_{n \to + \infty} u_{n} = + \infty o$ $s o t h e q u e s t i o n c o n t a i n a e r r o r . . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

	Answered by MJS last updated on 30/Apr/18

	${(\frac{{(\frac{n}{e})}^{n} \sqrt{2 π n}}{n^{n} m^{n}})}^{\frac{1}{n}} = \frac{{(2 π n)}^{\frac{1}{2 n}}}{e m} = \frac{{(2 π)}^{\frac{1}{2 n}}}{e m} \times n^{\frac{1}{2 n}}$ $\lim_{n \to \infty} \frac{{(2 π)}^{\frac{1}{2 n}}}{e m} \times n^{\frac{1}{2 n}} = \frac{1}{e m}$
	Commented by rahul 19 last updated on 30/Apr/18

	$m y d o u b t i s :$ $h o w \underset{n}{l} \underset{}{i} \underset{\infty}{m} n^{\frac{1}{2 n}} = 1 ?$
	Commented by MJS last updated on 30/Apr/18

	${\underset{n \to \infty}{\lim 10}}^{\log_{10} n^{\frac{1}{2 n}}} = \lim_{n \to \infty} 10^{\frac{\log_{10} n}{2 n}} = 1$ $f (n) = \frac{\log_{10} n}{2 n} (take any base > 1)$ $n = 10^{k}$ $f (10^{k}) = \frac{k}{2 \times 10^{k}}$ $\lim_{k \to \infty} \frac{k}{2 \times 10^{k}} = 0 \Rightarrow \lim_{n \to \infty} \frac{\log_{10} n}{2 n} = 0 \Rightarrow {\underset{n \to \infty}{\lim 10}}^{\frac{\log_{10} n}{2 n}} = 1$
	Commented by rahul 19 last updated on 30/Apr/18

	$T hank you sir!$
	Commented by prof Abdo imad last updated on 30/Apr/18

	$n^{\frac{1}{2 n}} = e^{\frac{1}{2 n} l n (n)} a n d w e k n o w t h a t {l i m}_{n \to + \infty} \frac{l n (n)}{n} = 0$ $\Rightarrow {l i m}_{n \to + \infty} n^{\frac{1}{2 n}} = e^{0} = 1 .$
	Commented by rahul 19 last updated on 01/May/18

	$T hank you sir!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum