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$n!\gt 2^{n}$ for $n \geq 4$.

Then there is an inductive proof:

We assume that $n! \gt 2^{n}$ for every random $n \in \mathbb{N}$ with $n\geq4$, and we need to proof that $(n+1)!\gt 2^{n+1}$. Because $(n+1)\gt0$ and $n!\gt2^{n}$, it follows that $(n+1)n!\gt(n+1)2^{n}$. In the left part of the equation we have $(n+1)!$, but in the right part we don't have $2^{n+1}$.

And here comes the part I don't understand:

We want to replace$(n+1)2^{n}$ with $2^{n+1}$, or $n+1$ by $2$. We can do this without any problems, because it is sure that for $n\geq4$ that $n+1\gt2$, and thus that $(n+1)\cdot2^{n}\gt2\cdot2^{n}$

How much I try, I just don't see the connection. So you multiply $n+1\gt2$ by $2^{n+1}$ on both sides, and that makes $(n+1)n!\gt(n+1)2^{n}$ into $(n+1)!\gt2^{n+1}$? I don't see how you can get from $(n+1)2^{n}$ to $2^{n+1}$ with what they're saying. To me, it seems like they're just saying "Hey, because $n+1\gt2$ is greater than $2$, we can just replace it with $2^{n+1}$".

Caleb Stanford
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Garth Marenghi
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  • If $f(n)=n!-2^n$

    we find $f(3)=-2<0, f(4)=4!-2^4=8>0$

    Let $f(n)>0$ for $n=m$

    So, $m!>2^m\implies (m+1)\cdot m!>(m+1)2^m>2\cdot2^m=2^{m+1}$

     – lab bhattacharjee Aug 25 '13 at 15:45
  • note $n \geq 4 \Rightarrow n \geq 2 \Rightarrow n 2^{n-1} \geq 2^n$ – AlexR Aug 25 '13 at 15:45
  • I fixed the text appearing twice, and also had one $2^{2}$ replaced at the beginning. – Garth Marenghi Aug 25 '13 at 15:50
  • Your logic at the very beginning needs to be refined. You wrote: "We assume that $n! > 2n$ for every random $n \in \mathbb N$ with $n \ge 4$." If you really do mean "for every random $n \in \mathbb N$", then you should explain that more clearly. But I think you mean "for some $n \in \mathbb N$". – Théophile Jan 07 '15 at 20:49

2 Answers2

1

You multiply $n!>2^n$ with $n+1$ and get $(n+1)!>(n+1)\cdot 2^n$. And you multiply $n+1>2$ with $2^n$ to get $(n+1)\cdot 2^n>2^{n+1}$. So all in all you have $(n+1)!>(n+1)\cdot 2^n>2^{n+1}$ as needed

Hagen von Eitzen
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1

Inductive works like this:

For $n=4$ we see: $4!=24>16=2^4$.

Now for $n=5$, $5! = 5\times 4!> 5\times 2^4>2\times 2^4=2^5$. The first inequality due to induction hypothesis. The second being just $5>2$ (and this is the part you didn't understand).

Try to work out detail argument for general $n\geq 4$.

Quang Hoang
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  • I get how $5!=5\cdot4!\gt5\cdot2^{4}$, and I get that in the part $5\cdot2^{4}\gt2\cdot2^{4}$ it is because $5\gt2$. But I'm still not clear as to why you append $\gt2\cdot2^{4}$ to $5!=5\cdot4!\gt5\cdot2^{4}$ – Garth Marenghi Aug 25 '13 at 16:05