How to find $\lim\limits_{n\to\infty}\frac{2^{2^n}}{n!}$?

Question

What I have done is $\frac{2^{2^n}}{n!}=\frac{2^{2^n}}{1 * 2 * \dots * n} \leq \frac{2^{2^n}}{n}$. Then I am stuck here. I wanted to do something like in this answer using Squeeze theorem. Is it right direction to proceed solving? If not, what can I do next?

Answer 1

Note that\begin{align}\frac{\frac{2^{2^{n+1}}}{(n+1)!}}{\frac{2^{2^n}}{n!}}&=\frac 1{n+1}2^{2^{n+1}-2^n}\\&=\frac{2^{2^n}}{n+1}.\end{align}Since$$\lim_{n\to\infty}\frac{2^{2^n}}{n+1}=\infty,$$you have$$\lim_{n\to\infty}\frac{2^{2^n}}{n!}=\infty.$$

Answer 2

You can show that $$\lim_{n\rightarrow\infty}\frac{2^{2^n}}{n!}=\infty$$ By showing that, for $n\geq 0$ $$2^{2^n}>n!$$

The induction step is as follows: $$\left(2^{2^n}\right)^2>2^{2^{n}}(n+1)>(n+1)n!=(n+1)!$$ Now we will be done if we show that $$2^{2^n}>n+1$$ This can also easily be done by induction, by noticing that $$2^{2^n}>n+1\\\Rightarrow 2^{2^{n+1}}=\left(2^{2^n}\right)^2>(n+2)n+1>n+2$$

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The reason, by the way, for why an exponential function grows faster, eventually, compared to a polynomial is because of l'Hôpital's rule. Let $e(x)$ be an exponential function, and $p(x)$ a polynomial of degree $n$. Then consider $$L=\lim_{x\rightarrow\infty}\frac{e(x)}{p(x)}=\pm\frac{\infty}{\infty}$$ Then l'Hôpital states that $$L=\frac{\frac{d}{dx}e(x)}{\frac{d}{dx}p(x)}=\dots=\lim_{x\rightarrow\infty}\frac{\frac{d^n}{dx^n}e(x)}{\frac{d^n}{dx^n}p(x)}$$ $\frac{d^n}{dx^n}p(x)$ is a constant, and $\frac{d^n}{dx^n}e(x)\rightarrow\infty$, so $L\rightarrow\infty$.

Answer 3

As an alternative, we have that

$$\frac{2^{2^n}}{n!}=e^{2^n\log 2-log (n!)}\to \infty$$

indeed by integral approximation

$$\log(n!)=\sum_{k=1}^{n} \log k \sim \int_1^n \log x dx =n\log n-n+1$$

and therefore

$$2^n\log 2-log (n!)\sim 2^n\log 2-n\log n=2^n\left(\log 2-\frac{n\log n}{2^n}\right)\to \infty $$