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Why is

$$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$

Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is

$$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$


This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

  • 8
    So $\frac{2^n}{n!}$ is always positive, right? If you can show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{2^n}{n!}$ is always so, then...2011-10-31
  • 3
    Thank you J.M., your solution was simple and worked well. I wish you had provided it in the form of an answer so that I could accept it!2011-10-31
  • 12
    Then the sequence converges, but not necessarily to zero.2011-10-31
  • 0
    Note that for $n \ge 4$, $n!=(6)(4\cdot 5\cdots n)$. But $4/2\ge 2$, $5/2 \ge 2$, and so on, so $\frac{2^n}{n!} \le \frac{8}{6}\frac{1}{2^{n-3}}$.2011-10-31
  • 1
    @JM: $0<\frac{1}{2}+2^{-(k+1)}<\frac{1}{2}+2^{-k}$, but that sequence does not converge to $0$.2011-10-31
  • 0
    Yeah, that bit I gave has to be finished by sandwiching or something.2011-10-31
  • 0
    @J.M. I have used sandwich theorem but you can see it's a little bit different2014-06-28

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