Is this a valid application of limit laws for sequences I am trying to show that the sequence $$\frac{2^{n}}{n!}$$ converges to 0, and I am wondering if one of my final steps is valid.
I first off proved by induction that $$0 \lt \frac{2^{n}}{n!} \le 2\frac{2}{3}^{n-2}$$
And then simply noted that since $\frac{2}{3} \lt 1$ , as n goes to infinity $\frac{2}{3}^{n-2}$ goes to zero ie its limit is zero and since 2 is a constant, it goes to 2. Then I used that $lim(xy)=lim(x)lim(y)$ to show that our sequence must be less than or equal to that of one which goes to zero and hence the limit is zero.
Does this make sense? Were there valid applications of the limit product law, etc?
Any corrections or suggestions? Thanks
 A: Remember to say that the second inequality sign in the displayed inequalities holds for $n \geq 3$; other things look fine to me.
We can make use of another observation too: note that if $n \geq 2$ then
$$
\frac{2^{n}}{n!} = \frac{2^{n}}{1\cdot 2 \cdots n} \leq \frac{2^{2}}{1\cdot 2}\cdot \frac{2}{n}.
$$
A: I'm assuming you mean $2\left(\frac{2}{3}\right)^{n-2}$ goes to zero as $n \to \infty$, and not $\frac{2}{3}2^{n-2}$ goes to zero as $n\to\infty$? If the former, then yes you could move to the limit law you mention. However, be careful simply stating that $\lim(xy) = \lim(x)\lim(y)$. I think what you are getting at is if $\lim(x)$ exists and $\lim(y)$ exists, then $\lim(x\cdot y)$ exists. Since you established the limit of $2$ is $2$ and the limit of $\left(\frac{2}{3}\right)^{n-2}$ is zero, then the limit of the product is indeed zero, forcing $\frac{2^n}{n!}$ to zero as you say.
A: Another way to show this is to write
$$\frac{2^n}{n!}=e^{n\log 2-\log n!}$$
Then, observe that
$$\begin{align}
\log n!&=\sum_{k=1}^n\log k\\\\
&=n\log n+\sum_{k=1}^n\log (k/n)\\\\
&>n\log n+n\int_0^1\log x\,dx\\\\
&=n\log n-n
\end{align}$$
and thus, for $n>2$
$$\begin{align}
n\log 2-\log n!&<n\log 2-n\log n+n\\\\
&=-n\left(\log (n/2)-1\right)\\\\
&\le -n\left(\frac{n-2}{n}\right)\\\\
&=-(n-2)
\end{align}$$
Finally, we have
$$\begin{align}
0&<\frac{2^n}{n!}\\\\
&=e^{n\log 2-\log n!}\\\\
&<\frac{e^2}{e^n}
\end{align}$$
and by applying the squeeze theorem, we have 
$$\lim_{n\to \infty}\frac{2^n}{n!}=0$$
as was to be shown!
