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? 
October 30th, 2016, 12:03 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614 
$\forall n>4,~s_n = \dfrac{2^n}{n!} < 2\cdot 1 \cdot \left(\dfrac 2 3\right)^{n2}$ $s_n>0$ $\displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n2}=0$ $0 < \displaystyle{\lim_{n\to \infty}} s_n < \displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n2}=0$ so $\displaystyle{\lim_{n\to \infty}} s_n =0$ by the squeeze theorem 
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! 
October 30th, 2016, 12:38 PM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614  
October 30th, 2016, 12:42 PM  #5 
Senior Member Joined: Feb 2014 Posts: 112 Thanks: 1  Yes and i have an answer, but i don't understand well.

October 30th, 2016, 12:48 PM  #6 
Senior Member Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614  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. 
October 30th, 2016, 03:36 PM  #7 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  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 "n1" 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 "n1" terms in each product? Thus $\displaystyle \begin{align*} n! > 2^{n1} \end{align*}$. So the denominator is much, much greater than the numerator, and thus the ratio $\displaystyle \begin{align*} \frac{2^{n1}}{n!} \end{align*}$ must get smaller and smaller. Thus $\displaystyle \begin{align*} \lim_{n \to \infty} \frac{2^{n1}}{n!} = 0 \end{align*}$, 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. 
October 30th, 2016, 04:56 PM  #8  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
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.  
October 30th, 2016, 05:45 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,700 Thanks: 2177 Math Focus: Mainly analysis and algebra 
The ratio test would be a lot easier.


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 07:05 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 04:34 AM 