Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can I solve this? Any hints?
 A: If we can use the fact that the series $\sum_{n=0}^\infty x^n/n!$ converges for each $x\in\mathbb R$, we can show that
$$
\frac{2^nn^2}{n!}=\frac{2^nn}{(n-1)!}=2^2\frac{2^{n-2}}{(n-2)!}\frac n{n-1}\to0
$$
as $n\to\infty$ since $2^{n-2}/(n-2)!\to0$ as $n\to\infty$.
A: $n!>2\cdot 3^{n-2}$ for $n\ge2$ is apparent from the product definition, so:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot 3^{n-2}}=\frac{9n^2}{4\left(\frac{3}{2}\right)^n}$$
Note that $\left(\frac{3}{2}\right)^n=\left(1+\frac{1}{2}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k\ge\binom{n}{3}\left(\frac{1}{2}\right)^3$, so we have:
$$\frac{2^nn^2}{n!}\le\frac{9n^2}{4\left(\frac{3}{2}\right)^n}\le\frac{9n^2}{4\binom{n}{3}\left(\frac{1}{2}\right)^3}=\frac{108n^2}{n(n-1)(n-2)}=\frac{108}{n}\frac{1}{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}\to0$$
A: Note that we can show that $2^nn^2 \le (n-1)!$ for $n \ge 10$ using induction. 
Then $$0 \le \lim_{n\to \infty}\frac{2^n n^2}{n!} \le \lim_{n\to \infty}\frac{(n-1)!}{n!}=0$$
A: If you have the Ratio Test at hand, let $a_n=2^nn^2/n!$ and compute
$${a_{n+1}\over a_n}={2(n+1)\over n^2}\to0$$
If you can't simply cite the Ratio Test, you can just show
$${a_{n+1}\over a_n}={2(n+1)\over n^2}\lt{1\over 2}\quad\text{for }n\ge5$$
(from $n^2-4n-4=(n-2)^2-8$) and thus
$$a_{5+n}\lt\left(1\over2\right)^na_5\to0$$
A: Note that we have $ 9! = 362880 > 262144 = 4^9 $, so that $ n! > 4^n $ for all $ n \geq 9 $ and we have
$$ 0 \leq \frac{2^n n^2}{n!} \leq \frac{n^2}{2^n} $$
We may evaluate $ \lim_{n\to \infty} n^2 2^{-n} = 0 $ using L'Hopital, and then the squeeze theorem yields the result.
A: Note that:
$$n!\ge1\cdot 2\cdot3\cdot3\cdots\cdot3\cdot(n-2)\cdot(n-2)\cdot(n-2)=2\cdot3^{n-5}(n-2)^3$$
So for $n>5$:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot3^{n-5}(n-2)^3}=\frac{3^5}{2}\left(\frac{2}{3}\right)^n\left(\frac{n}{n-2}\right)^3\frac{1}{n}<\frac{3^5}{2}\left(\frac{2}{3}\right)^5\left(\frac{5}{3}\right)^3\frac{1}{n}\to0$$
