Finding the limit of $F(n)=\frac{(-2)^{n} n!}{n^n}$ $\lim\limits_{n\to\infty}\frac{(-2)^{n}n!}{n^{n}}$
I think the answer is 0 and I tried to use the squeeze theorem.
$-(\frac{2}{n})^{n}\leq-\frac{2^{n}}{n^{n}}\leq-\frac{2^{n}n!}{n^{n}}\leq\frac{(-2)^{n}n!}{n^{n}}$
but I don't how to find an upper bound for this.
 A: The arithmetic mean of $2,\ldots,n-2$ is $\frac n2$, hence by AM-GM inequality (for $n>3$)
$$n!\le1\cdot\left(\frac n2\right)^{n-3}\cdot (n-1)\cdot n
<\frac{n^{n-1}}{2^{n-3}}$$
A: Alternative method:
It suffices to show that $|F(n)| \to 0$ as $n \to \infty$. For this, what you can do, is follow these steps:
(1) Prove that we can find an $N$ such that for $n \ge N$ we have $|F(n)| \ge |F(n+1)|$. So the tail of the sequence is decreasing.
(2) We also know that $|F(n)| \ge 0$. Bounded sequences that have a decreasing tail converge, so $L := \lim_\infty |F|$ exists.
(3) Knowing that $L$ exists and using $L = \lim_n |F(n)| = \lim_n |F(2n)|$, show that $L = 0$.
Good luck!
A: Integral comparison method (useful for monotonic functions like $\ln (n!)\,)$.
For $j\in\Bbb Z^+$ we have $\ln j=\int_j^{j+1}(\ln j).dt<\int_j^{j+1}\ln (t).dt.$
Therefore for  $2\le n\in\Bbb Z^+$ we have $$\ln(n!)=(\ln n)+\sum_{j=1}^{n-1}\ln j<$$ $$<(\ln n)+\sum_{j=1}^{n-1}\int_j^{j+1}(\ln t).dt=$$ $$=(\ln n)+\int_1^n(\ln t).dt=$$ $$=((n+1)\ln n)-n+1$$ and hence $$n!=\exp(\ln (n!))<\exp (((n+1)\ln n)-n+1)=\frac {en^{n+1}}{e^n}$$ and hence $$\left|\frac {(-2)^n n!}{n^n}\right|<\frac {2^n e n^{n+1}}{n^n e^n}=en\left(\frac {2}{e}\right)^n.$$
