Proving the Limit of n! / n^n = 0 to Using Sequence Definition

[courtrigrad]

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

 

[cepheid]

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 realize 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?[/itex]

 

[courtrigrad]

I don't think that's right

 

[Hurkyl]

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

 

[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).

 

[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...

 

[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.

 

[courtrigrad]

ok thanks a lot everyone. Thats what the book wanted me to do.