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Very simple question. I'm calculating the limit of $\frac{2^n}{n!}$ as $n \rightarrow \infty$. I've got a solution that I can't find anything wrong with, but I haven't seen it reproduced anywhere else and I'm wondering if I've missed something.

$$x_n=\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdots \frac{2}{n-1} \cdot \frac{2}{n} \implies \,\, 0 \le x_n \le\frac{4}{n}$$ and by Squeezing Theorem, since $\frac{4}{n} \rightarrow 0 $ as $n \rightarrow \infty$, then $x_n \rightarrow 0$ as $n \rightarrow \infty$.

I made the 'upper bound' $\frac{4}{n}$ because we have $\frac{2}{1} \cdot \frac{2}{2}$ at the start and $\frac{2}{n}$ at the end and everything in between is less than 1, so surely the greatest $x_n$ could ever be is $\frac{4}{n}$. Is this OK? Thank you.

EDIT: Ahh, wait. I think I see now; do we have to take $ 2 \cdot (\frac{2}{3})^{n-2}$ as the lower bound? I say this because by my previous argument, couldn't the greatest $x_n$ could ever be, be $\frac{4}{n-1}$, for example?

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    $\begingroup$ Looks good to me. $\endgroup$ – vadim123 Jan 9 '16 at 23:29
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    $\begingroup$ BTW similar estimate was employed in this answer. $\endgroup$ – Martin Sleziak Apr 17 '17 at 9:25
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Another fancier way to see that the answer is $0$ is the associated power series $\displaystyle \sum_{n=0}^{\infty} \dfrac{2^n}{n!}$ converges to $e^2$, thus its $n$-th term $\dfrac{2^n}{n!}$must converge to $0$

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In response to your EDIT:

If all you need to do is show that $\displaystyle\lim_{n \to \infty}\dfrac{2^n}{n!} = 0$, then using the bound $0 \le \dfrac{2^n}{n!} \le \dfrac{4}{n}$ for $n \ge 1$ and then applying the squeeze theorem is sufficient.

However, if the problem had instead asked you to show that $\displaystyle\sum_{n = 0}^{\infty}\dfrac{2^n}{n!}$ converges, then an upper bound of $\dfrac{4}{n}$ would not be sufficient. So in that case, you would want to use the bound $0 \le \dfrac{2^n}{n!} \le \dfrac{2}{3}\left(\dfrac{2}{3}\right)^{n-2}$ for $n \ge 2$ and then use the comparison test.

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You haven't missed anything, your solution is fine.

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By using Squeeze theorem, we show that $\frac{2^n}{n!}=0$.

$0\le\frac{2^n}{n!}=\frac{2*2*2*2....}{n(n-1)...3*2*1}$

$\frac{2*2*2*2....}{n(n-1)...3*2*!}\le\frac{2^3}{3!}.\frac{2}{n}$

$\lim_{n\to\infty}(\frac{2^3}{3!}.\frac{2}{n})=0$

Hence $\lim_{n\to\infty}\frac{2*2*2*2....}{n(n-1)...3*2*1)}=\lim_{n\to\infty}\frac{2^n}{n!}=0$

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