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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 $\lim_{n\to\infty}\frac{2^n}{n!} = 0$.

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up vote 1 down vote accepted

It looks good, but a few points:

Showing that $\lim_{n\rightarrow\infty}{2\over n}=0$ suffices (but you should use $4/n$, see below). This should be justified. To do so, you may use the squeeze theorem: $$ 0\le{2^n\over n!} ={2\over 1}\cdot \underbrace{ {2\over 2}\cdot{2\over 3 }\cdots{2\over n-1}}_{\le 1}\cdot{2\over n}\le {4\over n}. $$

Fifth sentence. You need to fix $\epsilon$ first. Start off by saying "Let $\epsilon>0$. Choose $N$ a positive integer so that $N>{4\over \epsilon}$. Then if $n>N$..."

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Given $\varepsilon>0$, you need to find $N$ large enough so that whenever $n>N$, then $$ \frac 21 \times \frac 22 \times \frac23\times\frac24\times\frac25\times\cdots\times \frac2n < \varepsilon. $$ But you've only found $N$ large enough to make $\dfrac2n<\varepsilon$. So you still have more to do.

What you've written after "in other words" does not express the same thing in different words. Rather, it's a different thing.

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A couple of things: first of all, writing $$\lim_{n \to \infty}\frac{2}{n} = 0 \ \ \Rightarrow \ \lim_{n\to\infty}\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n} = 0$$ isn't completely rigorous, unless you observe that all the factors before $\frac{2}{n}$ are $< 1$ (except for the first, which is $2$), which implies that their product is too. Thus you can write: $$0<\frac{2^2}{2!}\times\frac{2}{3}\times\frac{2}{4}\times\frac{2}{5}\times...\times\frac{2}{n}< 2 \cdot 1 \cdot \frac2n \rightarrow 0$$ because, as you have proved, $2/n \rightarrow 0$. Whithout this observation the statement isn't justified, since all the accumulating factors could have resulted in an unbounded quantity.
Also another minor consideration: in your proof that $\lim_{n \to \infty}\frac{2}{n} = 0$ your $N$ should be a natural number, such as $N = \left[\frac2{\epsilon} \right] + 1$.

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