Could someone please show me how to prove that this limit is 0 by the squeeze/sandwich theorem in a way as simple as possible. Im not very good with all the technical math descriptions. The sequence is a=(-4)^n/n!
Since $\displaystyle (-4)^n$ alternates between negative and positive, an obvious "sandwiching" will be between $\displaystyle \frac{4^n}{n!}$ and $\displaystyle -\frac{4^n}{n!}$. One way to prove that $\displaystyle \frac{4^n}{n!}$ goes to 0 is to show that $\displaystyle \sum \frac{4^n}{n!}$ converges since a necessary condition that a series converge is that the sequence of term go to 0. You can do that by using the "ratio test" or even by noting that the series converges to $\displaystyle e^4$.