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?
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Call that thing $a_n$. What can you say about $a_{n+1}/a_n$? – Angina Seng Sep 15 '18 at 17:11
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@LordSharktheUnknown, it is equal to $\frac{2^{2^n}}{n+1}$. Can't see anything special. What I can intuitively say is that $2^{2^n}$ grows much faster than $n+1$. What was I supposed to say then? – Turkhan Badalov Sep 15 '18 at 17:23
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See Jose's answer... – Angina Seng Sep 15 '18 at 17:24
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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$$\lim_{n\to\infty}\frac{2^{2^n}}{n!}=\infty.$$
José Carlos Santos
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How did you know $\displaystyle\lim_{n\to\infty}\frac{2^{2^n}}{n+1}=\infty$? – Turkhan Badalov Sep 15 '18 at 17:20
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Because if $a>1$, and $k\in\mathbb N$, you always have $\lim_{n\to\infty}\frac{a^n}{n^k}=\infty$. And $\frac{2^{2^n}}{n+1}>\frac{2^n}{n+1}$. – José Carlos Santos Sep 15 '18 at 17:24
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Wow, I have never seen this property before I think or maybe didn't have enough sleep :) Can I know the name of the property or where to read about it? – Turkhan Badalov Sep 15 '18 at 17:27
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@TurkhanBadalov I don't think it has a name. Peaple just say that an exponential always growths faster than a polynomial. – José Carlos Santos Sep 15 '18 at 17:32
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Can I ask how you made transition from $\displaystyle\lim_{n\to\infty}\frac{2^{2^n}}{n+1}=\infty$ to $\lim_{n\to\infty}\frac{2^{2^n}}{n!}=\infty$? In my understanding, you can do it if the last sequence is larger than the previous, so if the smaller one goes to infinity, the larger one also goes. But $\frac{2^{2^n}}{n!} \leq \frac{2^{2^n}}{n+1}$. What am I missing? – Turkhan Badalov Sep 15 '18 at 18:08
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Carlos.Please elaborate :You have a_{n+1}=[(2^(2^n))/(n+1)]a_n; Since the factor of a_n goes to infty you conclude that a_n goes to infty.Do I miss something, thanks, Peter – Peter Szilas Sep 15 '18 at 18:09
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@PeterSzilas Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\infty$, in particular $\frac{a_{n+1}}{a_n}\geqslant2$ if $n\geqslant N$, for some $N\in\mathbb N$. But then $a_{n+k}\geqslant2^ka_n$ and therefore $\lim_{n\to\infty}a_n=\infty$. – José Carlos Santos Sep 15 '18 at 18:12
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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 $$2^{2^n}>n+1\\\Rightarrow 2^{2^{n+1}}=\left(2^{2^n}\right)^2>(n+2)n+1>n+2$$
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 $L\rightarrow\infty$.
cansomeonehelpmeout
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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
$$2^n\log 2-log (n!)\sim 2^n\log 2-n\log n=2^n\left(\log 2-\frac{n\log n}{2^n}\right)\to \infty $$
user
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