I am trying to do an epsilonproof. My expression, let's name it $a(n)$, $a(n) = \frac{n!}{2^n}$ I have compared the ratio of $\frac{a(n+1)}{a(n)}$ , this shows me that $n!$ is growing faster than $2^n$. I want to prove that $\lim_{n\to\infty}a(n) = \infty$. If the numerator $n!$ is growing faster than the denominator $2^n$, I guess this must be true if $n\to \infty$. But the ratio only tells me that. I would also like to show that I can find, for all $\varepsilon > 0$, an $N$, s.t $n>N$ or $n=N$ which makes $a(n) > \varepsilon$. I have given a definition of an expression converging towards infinity and thus I need to show that according to my definition I can find an $N$ for all $\varepsilon$, s.t $\varepsilon < a(n)$. Here is where I have problem.
I can't get my head around what happens with $n!$ if its written in general form? I can't write it $\frac{n(n-1)}{ 2^n} > \varepsilon$, and then solve for $n$, since it is not a series of numbers. In programming I guess I could make an recursive expression for $n!$. Since I have the term $2^n$ in the denomenator I would assume that it would be some kind of logarithmic solution to find the calculation for $N$, with $\varepsilon$. Well..I hope someone can help me and set me going. This is an assignment for school and I really want to learn this and also get the assignment done. If I would have had $a(n) \to 1$ for $n\to\infty$. And $a(n) = \frac{n}{n+1}$ , it is much much easier. Then I would have written $1-\varepsilon < \frac{n}{n+1}$ and solved $n > 1/\varepsilon-1$. Now It's not that easy and I have no idea what to do with $\frac{n!}{ 2^n}$....please help. I think the biggest problem for me is the $n!$ term.
$$\frac{n!}{2^n}>\frac12\frac22\left(\frac32\right)^{n-2}>\epsilon$$ Can you solve the right-hand inequality for $\epsilon$ and $n$?