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!

Notice that
$$\frac{2^n}{n!} = {{2 *2*2*2*...*2} \over {1*2*3*4*...*n}}=2*{{2*2*...*2} \over {3*4*...*n}} \leq 2*1*1*...*{{2} \over {n}}={4 \over n}$$
So
$$0 \leq \frac{2^n}{n!} \leq {4 \over n}\xrightarrow[n\to\infty]{}0$$
And hence $\lim_{n \to \infty} \frac{2^n}{n!}=0$
