Find smallest n so that the finite sum $\displaystyle \sum_{k=1}^{n}\frac{(-2)^{k}}{k!}$ is certain to differ from $\displaystyle \sum_{k=1}^{\infty}\frac{(-2)^{k}}{k!}$ by less than .001.

So I set up the inequality:

$\displaystyle a_{n+1} \le .001$

$\displaystyle \frac{(-2)^{n+1}}{(n+1)n!} \le \frac{1}{100}$

$\displaystyle (n+1)n! \ge -200(-2)^{n}$

And that is where I am stuck. How do I solve for $\displaystyle n$?