real analysis - Prove that $frac{2^n}{n!}$ converges 0.

Prove that $\frac{2^n}{n!}$ converges 0.

I can see why, I just don't get how exactly to do convergence proofs. Right now I have:

For $n>6$, $|\frac{2^n}{n!}-0|=\frac{2^n}{n!}<\frac{2^n}{3^n}$

and

assuming $\frac{2^n}{3^n}<\epsilon$, $n<\frac{\ln\epsilon}{\ln\frac2 3}$

Not sure if the last step is even right...

(This was an exam question today)

Answer

I'm pretty sure that last one need to be $n > \frac{\ln \varepsilon}{\ln \frac{2}{3}}$. But then that this works. For every $\varepsilon$ you give an explicit way to find $N\left(= \frac{\ln \varepsilon}{\ln \frac{2}{3}})\right)$ such that for all $n > N$ we have $|x_n - 0| < \varepsilon$. Definitions, ta da!