How do I show that $2^n/n!$ (doesn't have to be 2) is bounded above? 
Possible Duplicate:
Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. 

Consider the sequence $(8^n/n!)n\in \mathbb{N}$. I am trying to show that it is bounded above (it is bounded below by 0) and applying the Squeeze's theorem to show that the limit is 0.
I thought about $2^n/n! < 2(2/n) < 2$
BUt the presence of the upperbound of 2 bothers me. 
 A: Let $a_n = \frac{8^n}{n!}$. Then for $n \ge 16$,  $a_{n+1} < a_n/2$. That's all you need to show that the limit is 0.
A: What about a proof using infinite series? Let $\,a>0\,$ and put
$$a_n:=\frac{a^n}{n!}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)!}\frac{n!}{a^n}=\frac{a}{n+1}\xrightarrow [n\to\infty]{} 0$$
So the ratio (D'Alembert's) test tells us the infinite positive series $\,\displaystyle{\sum_{n=1}^\infty a_n}\,$ converges and thus $\,\displaystyle{\lim_{n\to\infty} a_n=0}$ , and this is true for any $\,a>0\,$
A: Let $x_n=\frac{a^n}{n!}$ with $a>0$. then all $x_n$ are positive and we have $\frac{x_{n+1}}{x_n}=\frac{a}{n+1}<1$ for $n>a-1$, hence $x_{\lfloor a\rfloor}$ is an upper bound and from then on the sequence decreases monotonuously.
But you can get more:
$\frac{x_{n+1}}{x_n}=\frac{a}{n+1}<\frac 12$ for $n>2a-1$, hence $x_n< \frac C{2^n}$ for almost all $n$ for suitable $C$ (choose $C$ such the inequality holds for $n=\lfloor 2a\rfloor$).
A: $$0 \leq \frac{2^n}{n!} = \frac{2}{1} \frac{2}{2} \frac{2}{3} \cdots \frac{2}{n} = 2 \cdot \frac{2}{3} \cdots \frac{2}{n} \leq 2 \cdot \left(\frac{2}{3} \right)^{n-2}$$
and since $\frac{2}{3} < 1$, we know $\left(\frac{2}{3} \right)^{n-2} \to 0$ as $n \to \infty$.
For any number replacing 2, say $k$, eventually $n$ is bigger than $k$, so this same argument applies.
