Not sure where this should exactly go, but...

**Pinkk** · October 19th 2009, 04:47 PM

...For my proofwriting class, I have to show that $n! > a^{n}$ when $n \geq A$ , with $a,n \in \mathbb{N}$ . I know the eventual proof will be by induction, but I am not sure where to determine what the value of $A$ . Messing around with the numbers 1-9, I was able to generalize it to two cases: If $a$ is even, then $A = \frac{5a-2}{2}$ and if $a$ is odd, then $A = \frac{5a-1}{2}$ , but is there a way to generalize what $A$ must equal for ANY natural number $a$ ?

Edit: Looking at $a > 9$ , my generalization doesn't seem to hold. Now I am definitely confused.

Edit2: And just to clarify, I want $A$ to be the least possible value.

**Jameson** · October 26th 2009, 03:31 PM

I'm going to use the perks of my Admin status and bump this question. I'm stumped right now but I'm sure some of our other members could help you out.

EDIT: I've been thinking about this more and more and the only thing I could come with was to use Stirling's Approximation to help choose a generalized A.

$n! \approx \sqrt{2 \pi}n^{n+.5}e^{-n}$

**Pinkk** · October 26th 2009, 07:42 PM

Well, I went ahead with an induction proof using $A=\frac{5a-1}{2}$ when $A$ is odd and $A=\frac{5a-2}{2}$ when $A$ is even (I rechecked and those two cases of $A$ always hold). I showed it to my professor and she said it was good.

Someone also showed me a neat way of doing it for even numbers

$n \ge 2a^{2}$ where $n=2m$ , so $m \ge a^{2}$

It follows then that:

$n!=(2m)(2m-1)\cdot \cdot \cdot (m+1)(m)\cdot \cdot \cdot(2)(1) > (2m)(2m-1)\cdot \cdot \cdot(m+1)$ $> a^{2}\cdot \cdot \cdot a^{2} = (a^{2})^{m} = a^{2m} = a^{n}$

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Not sure where this should exactly go, but...