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For $n$ tending to infinity find the following limit $$2^n/n!$$ i have a feeling that it is multiplication of many numbers with the last one turning to 0 but the 1st one is finite so limit should be 0 .But i am not sure and neither am i able to put it in mathematical form.

Thank you for your help .

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Try the ratio test. –  aschepler Aug 9 at 15:05
    
ratio test ???? –  avz2611 Aug 9 at 15:06
    
    
ok got it , thank you for the help , –  avz2611 Aug 9 at 15:09
    
You can see that the factorial function grows much faster than the exponential function (meaning that it increases much faster), therefore the limit will converge to $0$. –  Vishwa Iyer Aug 10 at 0:42

6 Answers 6

up vote 12 down vote accepted

$$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$

As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$

Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$

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2  
Hey! I recognize this one!! –  I like Serena Aug 9 at 16:29

BIG HINT:

$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}=e^x<\infty$$

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You can use the theorem of d'Alembert for the sequences then you immediately have:

if $x_n=\frac{2^n}{n!}$,

$$\lim_{n\to\infty }\left|\frac{x_{n+1}}{x_n} \right|=\lim_{n\to\infty }\frac{2^{n+1}n!}{(n+1)! 2^n}=\lim_{n\to\infty }\frac{2}{(n+1)}=0<1$$

then $$\lim_{n\to\infty }\frac{2^n}{n!}=0.$$

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Hint: Use induction to prove $\frac{2^n}{n!}\le (\frac{1}{2})^{n-4}$ for $n\ge 4$.

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Hint: You can use Stirling's approximation $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n $$

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wow this is new to me , so i am going to like this one –  avz2611 Aug 9 at 15:11
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You're missing a power of $n$ on the RHS. –  Eul Can Aug 9 at 15:18
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Every time a factorial appears somewhere in a limit, someone recommends Stirling's approximation. –  Gahawar Aug 9 at 15:19
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Ridiculous hint. Verifying it is significantly harder than the question at hand. –  Andres Caicedo Aug 11 at 15:18
    
I don't think so, it is much simpler... –  MathGems Aug 15 at 12:16

One way to aproach these kinds of limits is to use the monotone convergence theorem, (real bounded monotone sequences converge). So for convergence you need to prove that 1. your sequence is monotone, 2. it's bounded

For your sequence you can prove that it is decreasing by using the ratio test as in idm's answer. And you can clearly see that it is bounded by 0. This means that a limit exists, let $a_n$ be your sequence, then

$$ a_{n+1} = \frac{2^{n+1}}{(n+1)!} = a_n\frac{2}{n+1} $$

Now because we know $\lim_{n \to \infty} a_n = a$, we can replace $a_n$ and $a_{n+1}$ in the above equation by their limit, when $n \to \infty$

$$ a = a(\lim_{n \to \infty}\frac{2}{n+1}) = 0 $$

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