How to prove convergence $a_n = n!/2^n$
Since $n!/2^n = 1/2 * 2/2 * 3/2 * ... * (n-1)/2 * n/2$
$1/2 <= n!/2^n$
I am having trouble finding an upper bound for the sequence.
Any hints or help is much appreciated in finding the upper bound of justifying why $a_n$ is unbounded?

Denote $f(n)=\frac{n!}{2^n}$
$f(4)=\frac{4!}{2^4}=\frac{3}{2}>1$ for 
$n\geq3$ we have $\frac{(n+1)!}{2^{n+1}}>2\frac{n!}{2^n}$.(since $n+1\geq4)$ 
So $f(n)\geq2f(n-1)$ for $n\geq 3$
since $f(4)\geq1$ we get $f(5)\geq2,f(6)\geq4,f(7)\geq 8\dots$
