Does n! grow faster than 2^n as n approaches infinity?

[Je m'appelle]

Homework Statement

Show that the following series diverges

$$\sum_{n=1}^{\infty}\frac{n!}{2^{n}}$$

Homework Equations

The Divergence Test: In order for a series to be divergent, the following must be true

$$\lim_{n\rightarrow \infty} a_n \neq 0$$, or

$$\lim_{n\rightarrow \infty} a_n \nexists$$

The Attempt at a Solution

Alright, I know how to work it out with the denominator, as it is a geometric series and therefore as $$n \rightarrow \infty,\ 2^{n} \rightarrow 1$$

But how do I do whenever I find a factorial? How do I work it out? I don't know what to do with this factorial, can I assume the following in this case

As $$n \rightarrow \infty$$,

$$n! \approx n$$

Then as $$n \rightarrow \infty$$ it would summarize to

$$a_n = \frac{n}{2^{n}}$$, so by using L'hôpital's

$$\frac{\frac{d}{dn}n}{\frac{d}{dn} 2^{n}}$$

$$\frac{1}{2^{n}}$$, and then as $$n \rightarrow \infty$$,

$$\frac{1}{1} = 1 \neq 0$$Is this it?

 

[╔(σ_σ)╝]

No.
n!=n(n-1)(n-2)... *3*(2)*(1) [ n terms]
$$2^{n}= 2*2*... *2$$ n times.

I have no idea what you are doing there.

Please write the expansion of n! and $$2^{n}$$ as the numerator and denominator then try to see why the limit does not exist.

 

[Je m'appelle]

Nevermind what I wrote before.

I was trying to show that the series was divergent by $$\lim_{n\rightarrow \infty} a_n \neq 0$$, but that's not possible because the limit doesn't exist, that's why I messed up. But by using $$\lim_{n\rightarrow \infty} a_n \nexists$$ it can be shown that the series diverge. I had forgotten this second case and that's why I did some mess before on working out the first case heh.