It's been a couple years since I've done analysis, so I was hoping someone could point out any possible flaws I have in the following proof.
Prove $\lim_{n\to\infty}\frac{2^n}{n!} = 0$.
For $n > 2$, we have: $\lim_{n\to\infty}\frac{2^n}{n!} = \lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times\cdots\times\frac{2}{n}$.
Need to show that $\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times\cdots \to 0$. In other words, need to show that $\lim_{n\to\infty}\frac{2}{n} = 0$:
$\forall\epsilon\in\mathbb{N}^+, \exists N\in\mathbb{N} $ such that $\forall n>N$ we have $\frac{2}{n} < \epsilon$. Take $N = \frac{2}{\epsilon}$. Since $n>N$ we have $n > \frac{2}{\epsilon}$. Thus $\frac{2}{n} < \epsilon$. Therefore $\lim_{n\to\infty}\frac{2}{n} = 0$.
And so $\lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n} = 0$.
Then $\lim_{n\to\infty}\frac{2^n}{n!} = 0$.