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Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.
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)