Answer
The sequence diverges.
Work Step by Step
By definition
$n!=1 \times 2 \times 3... \times (n-1) \times n$
$\lim\limits_{n \to \infty} \frac{n!}{2^{n}}$
$=\lim\limits_{n \to \infty} \frac{1 \times 2 \times 3... \times (n-1) \times n}{2 \times 2 \times 2... \times 2 \times 2}$
$=\lim\limits_{n \to \infty} \frac{1}{2} \times \frac{2}{2} \times \frac{3}{2}... \times \frac{n-1}{2} \times \frac{n}{2}$
$=\lim\limits_{n \to \infty} \frac{1}{2} \times \frac{2}{2} \times [\frac{3}{2}... \times \frac{n-1}{2} \times \frac{n}{2}]$
The part in brackets is a product of infinity and greater than 1, therefore the part in brackets equals to $\infty$
$=\lim\limits_{n \to \infty} \frac{1}{2} \times \frac{2}{2} \times \infty = \infty$
The sequence diverges.
