How to prove that $n^2(2^n) / n! \to 0$ as $n \to \infty$ I'm trying to see how I can prove that :
$$\frac{n^22^n}{n!}\xrightarrow [n\to\infty]{} 0$$
Not sure how to show this..anyone care to explain?  Thanks!
 A: A fancy way: check for convergence/divergence of the series $\,\displaystyle{\sum_{n=1}^\infty\frac{n^22^n}{n!}}\,$ , for example with the ratio test (D'Alembert's test):
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^22^{n+1}}{(n+1)!}\frac{n!}{n^22^n}=2\frac{n+1}{n^2}\xrightarrow[n\to\infty]{} 0<1$$
Thus, the ratio test tells us the series converges and so its general term sequence converges to zero.
A: Here's a hint:
$$
\begin{align}
\frac{n^2 2^n}{n!}&=\frac{\overbrace{2\times 2 \times \cdots \times 2}^{n \mbox{ times}} \times n^2}{1\times 2\times \cdots \times 3\times n}\\
&=\frac{n}{n-1} \times 4 \cdot \frac{\overbrace{2\times 2\times \cdots \times 2}^{(n-2) \mbox{ times}}}{1\times 2\times \cdots \times (n-2)}\\
\end{align}
$$
If you still need help, please indicate so in the comments.
A: By Stirling's Approximation:
$$\frac{n^2 2^n}{n!} \to \frac{(2e)^n}{\sqrt{2\pi}n^{n-1.5}}\sim \frac{1}{(n/2e)^n}\to 0$$
A: Let $$a_n=\frac{n^22^n}{n!}$$ for all $n\ge1$. Note that each $a_n$ is positive, and that $$\begin{align}\frac{a_{n+1}}{a_n} &=\cfrac{\left(\frac{(n+1)^22^{n+1}}{(n+1)!}\right)}{\left(\frac{n^22^n}{n!}\right)}\\ &= \frac{(n+1)^2}{n^2}\cdot\frac{2^{n+1}}{2^n}\cdot\frac{n!}{(n+1)!}\\ &= \frac{(n+1)^2}{n^2}\cdot2\cdot\frac{1}{n+1}\\ &= \frac{2n+2}{n^2}\\ &\leq\frac{4n}{n^2}\\ &=\frac4n\end{align}$$ for each $n$. In particular, then, $$a_{n+1}\leq \frac45a_n$$ for all $n\geq 5$, so that $$a_n\leq\left(\frac45\right)^{n-5}a_5$$ for all $n\geq 5$ by simple induction. Now use the Squeeze Theorem.
A: Use the squeeze theorem. You know that it is $\geq 0$ for each $n$. Now, by looking at the independent factors try to find an upper bound that goes to $0$:
$$\frac{n^22^n}{n!}=2^3\frac{n2^{n-3}}{(n-1)!}$$
We have $n-2$ factors both in the numerator and denominator (I omit the $1$ in the factorial)
This gets us, after $n=4$, to
$$\eqalign{
  & {{4 \cdot {2^1}} \over {3!}} = {4 \over 3}{2 \over 2} \le {4 \over 3}{2 \over 2}  \cr 
  & {{5 \cdot {2^2}} \over {4!}} = {5 \over 4}{2 \over 3}{2 \over 2} \le {5 \over 4}{2 \over 3}  \cr 
  & {{6 \cdot {2^3}} \over {5!}} = {6 \over 5}{2 \over 4}{2 \over 3}{2 \over 2} \le {6 \over 5}{2 \over 4} \cr} $$
So in general, we can see
$$a_n\leq \frac{16n}{(n-2)(n-1)}$$
for $n\geq 4$. This means that $a_n\to 0$, by the squeeze theorem.
