# Proof of limit

1. Nov 15, 2004

how would you prove

lim (n! / n^n) = 0
n--> 00

Should I use the definition of a sequence? like n! < what

n^n is less than what? and find limit of this?

thanks

2. Nov 15, 2004

### cepheid

Staff Emeritus
I'm sorry...I'm kind of rusty with this stuff, so I can't offer anything that would constitute a proof, but at first glance, I noticed the following:

$$n^n = n \cdot n \cdot n \cdot n...n \cdot n \cdot n$$
'n' times

WHEREAS

$$n! = n(n-1)(n-2)...3 \cdot 2 \cdot 1$$.

Clearly, the first product is larger than the second, and in the limit as n approaches infinity, the difference between them will be on the order of infinity as well, so if the denominator is infinitely larger than the numerator...

I realise this doesn't help answer the question. Also, I've heard that it is not correct to think of a power such as [itex] x^n [/tex] as 'n' x's multiplied together, but I cannot remember why. Can someone clarify?

Last edited: Nov 15, 2004
3. Nov 15, 2004

I dont think thats right

4. Nov 15, 2004

### Hurkyl

Staff Emeritus
Maybe you can split the fraction into the product two parts... one that goes to zero, and one that's bounded...

5. Nov 15, 2004

### AKG

Just a rough guess:

Use the ratio test to show that the sequence converges (where n only takes on integral values) and then use that to show that the function converges (assuming there some sort of rule that says you can do that).

6. Nov 16, 2004

### Galileo

You could show that the sequence is monotonically decreasing.
i.e. show that:
$$\frac{(n+1)!}{(n+1)^{n+1}}<\frac{n!}{n^n}$$
Together with the fact that the sequence is bounded below...

7. Nov 16, 2004

### arildno

The simplest way is to look at the sequence using even integers (n=2m):
Then:
$$n!\leq(\frac{n}{2})^{\frac{n}{2}}n^{\frac{n}{2}}$$
$$n^{n}=n^{\frac{n}{2}}n^{\frac{n}{2}}$$
Hence, your ratio is bounded by:
$$\frac{n!}{n^{n}}\leq(\frac{1}{2})^{\frac{n}{2}}$$

This is, of course, along the lines Hurkyl indicated.

Last edited: Nov 16, 2004
8. Nov 16, 2004