October 30th, 2016, 11:10 AM  #1 
Senior Member Joined: Feb 2014 Posts: 110 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: 603 Thanks: 324 
$\forall n>4,~s_n = \dfrac{2^n}{n!} < 2\cdot 1 \cdot \left(\dfrac 2 3\right)^{n2}$ $s_n>0$ $\displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n2}=0$ $0 < \displaystyle{\lim_{n\to \infty}} s_n < \displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n2}=0$ 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: 110 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: 603 Thanks: 324  
October 30th, 2016, 12:42 PM  #5 
Senior Member Joined: Feb 2014 Posts: 110 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: 603 Thanks: 324  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: 38 Thanks: 19  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: 6,346 Thanks: 2084 Math Focus: Mainly analysis and algebra 
The ratio test would be a lot easier.


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