My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Thanks Tree5Thanks
  • 1 Post By romsek
  • 1 Post By romsek
  • 3 Post By Prove It
Reply
 
LinkBack Thread Tools Display Modes
October 30th, 2016, 11:10 AM   #1
Senior Member
 
Joined: Feb 2014

Posts: 112
Thanks: 1

Limit of Fraction

Hello,
What is limit of Fraction?
Attached Images
File Type: jpg zas.JPG (9.1 KB, 17 views)
life24 is offline  
 
October 30th, 2016, 12:03 PM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,410
Thanks: 715

$\forall n>4,~s_n = \dfrac{2^n}{n!} < 2\cdot 1 \cdot \left(\dfrac 2 3\right)^{n-2}$

$s_n>0$

$\displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n-2}=0$

$0 < \displaystyle{\lim_{n\to \infty}} s_n < \displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n-2}=0$

so $\displaystyle{\lim_{n\to \infty}} s_n =0$ by the squeeze theorem
romsek is online now  
October 30th, 2016, 12:31 PM   #3
Senior Member
 
Joined: Feb 2014

Posts: 112
Thanks: 1

Thank you,
I got confused
I want solve this limit by Stirling's approximation for n!
life24 is offline  
October 30th, 2016, 12:38 PM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,410
Thanks: 715

Quote:
Originally Posted by life24 View Post
Thank you,
I got confused
I want solve this limit by Stirling's approximation for n!
well do you know what Stirling's approximation is?
Thanks from life24
romsek is online now  
October 30th, 2016, 12:42 PM   #5
Senior Member
 
Joined: Feb 2014

Posts: 112
Thanks: 1

Quote:
Originally Posted by romsek View Post
well do you know what Stirling's approximation is?
Yes and i have an answer, but i don't understand well.
Attached Images
File Type: jpg bin.JPG (19.5 KB, 10 views)
life24 is offline  
October 30th, 2016, 12:48 PM   #6
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,410
Thanks: 715

Quote:
Originally Posted by life24 View Post
Yes and i have an answer, but i don't understand well.
I don't think you need to understand how it's derived. Just understand how to use it.

The limit in your picture is the reciprocal of the limit you want.

The numerator of the 2nd term is Stirling's approximation of $n!$

the rest is just algebra.
Thanks from life24
romsek is online now  
October 30th, 2016, 03:36 PM   #7
Member
 
Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
Originally Posted by life24 View Post
Hello,
What is limit of Fraction?
Notice that $\displaystyle \begin{align*} n! = n \cdot \left( n - 1 \right) \cdot \left( n - 2 \right) \cdot \left( n - 3 \right) \cdot \dots \cdot 3 \cdot 2 \end{align*}$ (so exactly "n-1" terms in the product).

Can you see that

$\displaystyle \begin{align*} 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot \left( n - 2 \right) \cdot \left( n - 1 \right) \cdot n > 2 \cdot 2 \cdot 2 \cdot 2 \cdot \dots \cdot 2 \cdot 2 \cdot 2 \end{align*}$

when there are exactly "n-1" terms in each product?

Thus $\displaystyle \begin{align*} n! > 2^{n-1} \end{align*}$. So the denominator is much, much greater than the numerator, and thus the ratio $\displaystyle \begin{align*} \frac{2^{n-1}}{n!} \end{align*}$ must get smaller and smaller.

Thus $\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2^{n-1}}{n!} = 0 \end{align*}$, and multiplying by an extra 2 is not going to change th at. So $\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2^n}{n!} = 0 \end{align*}$ also.
Thanks from greg1313, topsquark and life24
Prove It is offline  
October 30th, 2016, 04:56 PM   #8
Banned Camp
 
Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

Quote:
Originally Posted by Prove It View Post


Thus $\displaystyle \begin{align*} n! > 2^{n-1} \end{align*}$. So the denominator is much, much greater than the numerator,
and thus the ratio $\displaystyle \begin{align*} \frac{2^{n-1}}{n!} \end{align*}$ must get smaller and smaller.

Thus $\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2^{n-1}}{n!} = 0, \end{align*} \ \ \ \ \ \ \ \ $That doesn't follow, because the product of the ratios could be
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $approaching a relatively small, but nonzero number instead.

and multiplying by an extra 2 is not going to change that. So $\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2^n}{n!} = 0 \end{align*}$ also.

I'm stating there is something you are not addressing in your informal proof.

.

Last edited by Math Message Board tutor; October 30th, 2016 at 04:58 PM.
Math Message Board tutor is offline  
October 30th, 2016, 05:45 PM   #9
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,939
Thanks: 2266

Math Focus: Mainly analysis and algebra
The ratio test would be a lot easier.
v8archie is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

Tags
fraction, limit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How to simplify a fraction times by a fraction Sammy Elementary Math 2 August 15th, 2014 02:46 AM
fraction in negative exponentiation fraction ThePope Elementary Math 3 November 4th, 2012 06:05 AM
How can I move a fraction to the bottom of a fraction daigo Algebra 3 July 15th, 2012 01:05 AM
Basic Algebra, fraction of a fraction... Tommy_Gun Algebra 5 June 3rd, 2012 06:14 PM
Fraction and Fraction Sum Substitution K Sengupta Elementary Math 1 February 9th, 2010 03:34 AM





Copyright © 2017 My Math Forum. All rights reserved.