# Thread: Not sure where this should exactly go, but...

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

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

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

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

2. 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.

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

3. Well, I went ahead with an induction proof using $\displaystyle A=\frac{5a-1}{2}$ when $\displaystyle A$ is odd and $\displaystyle A=\frac{5a-2}{2}$ when $\displaystyle A$ is even (I rechecked and those two cases of $\displaystyle A$ 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

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

It follows then that:

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