# Lightning question before the test : )

1. Nov 6, 2007

### frasifrasi

How can I show that the sequence { 4^(n)/n! }

converges to zero (by squeeze theorem)??

Thank you! : )

2. Nov 6, 2007

### dynamicsolo

If the Squeeze Theorem is required as part of the proof, how about comparing

$$\frac{4^{n}}{n!}$$ < $$\frac{4^{n}}{n^{n}}$$ (for n > 1)

and showing that the latter goes to zero as n--> infinity?

3. Nov 7, 2007

### Gib Z

Minor correction, but the "for n > 1" part isn't correct. You can find the correct value, or just say sufficiently large n.

4. Nov 7, 2007

### learningphysics

as n->infinity n^n>n! so $$\frac{4^n}{n!} > \frac{4^{n}}{n^{n}}$$.

I think probably the idea being expressed was to use a constant in the denominator instead of n. $$\frac{4^n}{8^n}$$. (or any constant^n in the denominator where the constant>4)

$$\frac{4^n}{n!}< \frac{4^n}{8^n}$$ for large enough n.

Last edited: Nov 7, 2007
5. Nov 7, 2007

### dynamicsolo

Whoops! Quite so -- I was only looking at the denominators. That needs to be n > 2 , doesn't it? (n=3 works...)