Why does this Calculus II sequence diverge?

Question

What steps should I take to understand why this sequence diverges (specifically, not to +∞ or -∞)?

$$\lim_{n\to\infty}\frac{8(n!)}{(-2)^n}$$

Answer 1

1. Show that the absolute values diverges to $+\infty$. (Hint: $n! > 2\cdot 3^{n-2}$ for $n \geq 3$.)
2. Show that the series alternates, i.e. $a_{2n} > 0$ while $a_{2n+1} < 0$.
3. Conclude that the series does not converges (from 1.), and does not diverge to $+\infty$ nor to $-\infty$ (from 2.).

Answer 2

$n!$ "grows" faster then $2^n$. so the $$\frac{8(n!)}{(-2)^n}$$ approaches $\infty$ faster then it approaches zero. as $n\rightarrow\infty$ also $(-2)^n$ oscillates between positive and negative values.

Answer 3

Hint: Because the terms of the sequence alternate in sign, the sequence can converge if and only if it converges to zero.

And the magnitude of the $n^{\textrm{th}}$ term is $$8\cdot\frac{1}{2}\cdot\frac{2}{2}\cdot \frac{3}{2}\cdots\frac{n}{2}$$ Note that all factors beyond the third are larger than $1$, so do you see why the terms can't converge to zero?