Convergence, finding limit $\lim_{n \to \infty} \frac{2^n}{n!}$ I just came across an exercise, however I don't know how to find the limit of $$\lim_{n \to \infty} \frac{2^n}{n!}$$ can any body help? 
Of course this is not homework, I'm only trying out example myself, from https://people.math.osu.edu/fowler.291/sequences-and-series.pdf page 38.
I know that the limit exists, and it is $0$, checked it in on the wolfram, but I don't know how to solve it.
thanks in advance!

Generalization for $1<a<k<n$: 
$$
\lim_{n \to \infty} \frac{a^n}{n!} < a^{k-1} \lim_{n \to \infty} \frac{a^{n-k}}{k(k+1) \ldots n} = k^k \lim_{n \to \infty} \big(\frac{a}{k})^n =0
$$
The last step is since $k(k+1) \ldots n >k^{n-k}$
