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I am stuck at one problem. So I have to check if sequence is convergent.

$$\frac{2^x}{x!}$$

My thinking was to calculate limit and if limit exists it's convergent, but I am struggling with this:

$$\lim_{x \to \infty} \frac{2^x}{x!} = ?$$

I can't do it with L'Hopital's rule or with logarithm. Does that mean that this sequence is divergent?

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You could note that for $n\ge 4$ you have $$\frac{2^n}{n!} = \frac2n \frac{2^{n-1}}{(n-1)!} \le \frac12 \frac{2^{n-1}}{(n-1)!}.$$ Thus by induction you get that for the $n$-th term $$\frac{2^n}{n!}\le \frac{1}{2^{n-3}} \frac{2^3}{3!}=2^{-n} \ 64/6.$$ And, I assume you can finish this.

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  • $\begingroup$ I will try that way. Will get back here soon. $\endgroup$ Oct 20 '14 at 11:48
  • $\begingroup$ I am still struggling with this :). So I have one more question. Can I take any $$k$$ and prove that for this $$n>k$$ sequence converges and that means that whole sequence converges? $\endgroup$ Oct 21 '14 at 11:39
  • $\begingroup$ This is not what I meant. Let us set $a_n = 2^n/n!$. The above shows $a_n \le 2^{-n} (64/6)$. Also $0 \le a_n$. If you can show that $2^{-n} (64/6)$ tends to $0$ as $n$ to $\infty$ you have showed that the original sequence converges to $0$. And showing $2^{-n}$ goes to $0$ is likely somenting you saw already. $\endgroup$
    – quid
    Oct 21 '14 at 12:30
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We know that$$e^x=\sum\dfrac{x^n}{n!}.$$ So we have $$e^2=\sum\frac{2^n}{n!}\implies \lim_{n\to\infty}\frac{2^n}{n!}=0$$

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    $\begingroup$ I'm not sure he can use this, or else he wouldn't be asking this question $\endgroup$
    – mvggz
    Oct 20 '14 at 11:46
  • $\begingroup$ I didn't now for first expression. I will have to check it I guess! But this looks really nice! $\endgroup$ Oct 20 '14 at 11:49
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I take it from your notation that x is a positive integer. let $U_x$ be your sequence.

Then suppose x>3 :

x!=1*2*3*..*x > $2*3^{x-2}$

-> 0 < $U_x < \frac{2^x}{2*3^{x-2}} = 3*(\frac{2}{3})^{x-1}$

=> ($U_x$) ->0 , when x->∞

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  • $\begingroup$ But what if x is number in R? Because x can be negative too. $\endgroup$ Oct 20 '14 at 11:50
  • $\begingroup$ No it cannot be because x! doesn't exist for a negative integer, or at least should you prove very carefully what you mean by x! Which brings me to the next point: x! only exists for positive integers, when x is just a real number you want to study the gamma function evaluated in x+1. Doing so you can frame it with factorials of floor(x) and floor(x+1). Here is the wiki link: en.wikipedia.org/wiki/Gamma_function $\endgroup$
    – mvggz
    Oct 20 '14 at 11:54

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