Show that $ \lim_{n \rightarrow \infty} \frac{n!}{2^{n}} = \infty $ Show that $ \lim_{n \rightarrow \infty}  \frac{n!}{2^{n}} = \infty $
I know what happens intuitively.... 
$n!$ grows a lot faster than $2^{n}$ which implies that the limit goes to infinity, but that's not the focus here.
I'm asked to show this algebraically and use the definition for a limit of a sequence.
"Given an $\epsilon>0$ , how large must $n$ be in order for $\frac{n!}{2^{n}}$ to be greater than this $\epsilon$ ?"
My teacher recommends using an inequality to prove it but I'm feeling completely lost...
 A: For $n\geq8$
$$\frac{n!}{2^n}=\frac{2\cdot3}{2^4}\frac{4\cdot5...(n-1)}{2^{n-4}}n>\frac{2\cdot3}{2^4}\frac{4^{n-4}}{2^{n-4}}n=\frac{3}{2^4}2^{n-4}n\geq n$$
So, $n>\max(8,\epsilon)$ works.
A: $$
\frac{a_{n+1}}{a_n} = \frac{n+1}{2} \\\lim_{n\to \infty}  \frac{n+1}{2} \to \infty
$$
A: Let's call $u_n=\dfrac{n!}{2^n}$
$\dfrac{u_{n+1}}{u_n}=\dfrac{n+1}{2}$
A: Using Taylor series $ e^n = \frac1{0!}+\frac n{1!}+...+\frac{n^n}{n!}$ gives $n!>\frac{n^n}{e^n}$. So, $\frac{n!}{2^n} > \frac{n^n}{e^n2^n}=(\frac{n}{2e})^n$. We observe that RHS$ \to \infty $   as $n\to \infty$
A: If you need to do it by definition then you have to show that $\forall M>0 : \exists N(M) \in \mathbb N,  \exists n > N : |a_n| > M$, or in this particular case $|\frac{n!}{2^n}| > M$. The answer should be function $N(M)$. As $\frac{n!}{2^n} > 0$, we can write $\frac{n!}{2^n} > M$, or $n! > M*2^n$. If we take $N = \lceil M \rceil$ and $n = N+3$, then $(\lceil M \rceil+3)! > M*2^n$, or $1 * 2 * 3 * (\lceil M \rceil + 3) > 2 * 2 * 2 * \dots * 2 * M$.
This is true because we have at least $4$ elements in LHS and for that case this inequality is true: $1 * 2 * 3 * 4 > 2 * 2 * 2 * 2 * M, M \in (0,1]$ and $\forall M > 1$ we will multiply the left part by $\lceil M \rceil + 3$ (which is at least $=5$) while RHS only by $2$.
