# What n is required to make n! > a^n

1. May 2, 2007

### uart

I'm looking for a bound on n ( that is, "n" as a function of "a") to satisfy,

$$n! \geq a^n$$

It doesn't have to be the lowest possible bound, preferably one that's relatively easy to prove.

Thanks.

2. May 2, 2007

### uart

OK I've just found Sterling's formula, looking at that I can deduce that $$n = \lceil e\, a \rceil$$ will definitely work.

Does anyone have a bound that is very easy proof. Maybe something like n = 3a or even larger.

Last edited: May 2, 2007
3. May 2, 2007

### Office_Shredder

Staff Emeritus
Well, up until n=a, you clearly can't get anywhere (and have a-1 more a's on the RHS than the LHS)

Then you add up an a (actually, a number greater than a against a)for every one added to n until you hit a2

Then you have a-2 a's on the RHS compared to the LHS. From then on, every time you add one to n, you multiply the LHS by something greater than a2, and the RHS by a. So if n=a2+a (so you pick up a extra a's from the a2 and larger terms to mitigate the a-1 you felll behind by at the start), then n! >= an

That should work. If it needs clarification, I'll try to write it out a bit neater

4. May 2, 2007

### uart

Ok good, going as high as sqaures makes it pretty easy.

Take n = a^2 (for a>=2). You can just compare the terms (in the expansion of LHS and RHS) in pairs and show that every chosen pair-product on the LHS is greater than or equal to that of each RHS pair (which is of course always a*a on the RHS).

eg

1 * a^2 = a*a
2 * (a^2-1) > a*a
3 * (a^2-2) > a*a
....
k ^ (a^2+1-k) > a*a : { down to k = floor(a^2 / 2) }

That's the type of thing I was looking for, something that's like a "handwaving + by inspection" type of "proof". ;)

Last edited: May 2, 2007