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October 30th, 2016, 12:10 PM   #1
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Limit of Fraction

Hello,
What is limit of Fraction?
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 October 30th, 2016, 01:03 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,694 Thanks: 860 $\forall n>4,~s_n = \dfrac{2^n}{n!} < 2\cdot 1 \cdot \left(\dfrac 2 3\right)^{n-2}$ $s_n>0$ $\displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n-2}=0$ $0 < \displaystyle{\lim_{n\to \infty}} s_n < \displaystyle{\lim_{n\to \infty}}2\left(\dfrac 2 3\right)^{n-2}=0$ so $\displaystyle{\lim_{n\to \infty}} s_n =0$ by the squeeze theorem
 October 30th, 2016, 01:31 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, 01:38 PM   #4
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Quote:
 Originally Posted by life24 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?

October 30th, 2016, 01:42 PM   #5
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Quote:
 Originally Posted by romsek well do you know what Stirling's approximation is?
Yes and i have an answer, but i don't understand well.
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October 30th, 2016, 01:48 PM   #6
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Quote:
 Originally Posted by life24 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.

October 30th, 2016, 04:36 PM   #7
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Quote:
 Originally Posted by life24 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.

October 30th, 2016, 05:56 PM   #8
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Quote:
 Originally Posted by Prove It 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 05:58 PM.

 October 30th, 2016, 06:45 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra The ratio test would be a lot easier.

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