Find the limit of the sequence $(4^n)/((2n)!)$ How to find 
$$\displaystyle \lim_{n\to\infty} \frac{4^n}{(2n)!}$$
Can I use l'Hopital's rule?
 A: I don't see an application of l'hopital's rule here but using Stirling approximation you can deduce that the limit is zero.  
If you don't want to use any advanced results like Stirling, you can always go for an elementary proof like the following 
$$\frac{4^n}{(2n)!} \leq \frac{4^n}{(n+1)(n+2)...(2n)} \leq \frac{4^n}{(n+1)^n} $$
for $ n \geq 4$ we have  $$ \frac{4^n}{(2n)!} \leq \frac{4^n}{(n+1)^n} \leq (\frac{4}{5})^n $$  and then by taking the limit we find $0$.
A: $$\sum_{n\geq 0}\frac{4^n}{(2n)!} = \cosh(2)$$
hence the limit is trivially zero.
A: Here is another way to do it, if you really want to use Hopital (which I don't recommend) consider
$$\lim_{x \to \infty}\frac {4^x}{\Gamma(2x+1)}$$
Now you can use Hopital! :-P 
Suppose the above limit exists (call it $l$) and use hopital to find that 
$$l = \lim_{x \to \infty}\frac {4^x}{\Gamma(2x+1)} = \lim_{x \to \infty}\frac {4^x}{\Gamma(2x+1)} \cdot \lim_{x \to \infty}\frac{\ln 4}{2\psi(2x+1)} = l \cdot 0 = 0$$ and hence $l = 0$ . (here $\psi$ is the digamma function, and we know that $\lim_{x\to \infty} \psi(x) = \infty$. Also $\Gamma'(x) = \Gamma(x) \psi(x)$  )
Now since if the limit exists it must be the same on any subsequence. So set $x = n \in \mathbb N$ and the limit will be $0$. But setting $x = n$ you find 
$$\lim_{n \to \infty} \frac{4^n}{\Gamma(2n+1)} = \frac{4^n}{(2n)!} = 0$$
A: Here is one way.  Call the term $a_n$.  Then
$$a_n=\frac4{1\times2}\frac4{3\times4}\frac4{5\times6}\cdots\frac4{(2n-1)(2n)}\ .$$
Now it is easy to see that every term except the first is less than $\frac12$, so
$$0<a_n<\frac2{2^{n-1}}\ .$$
The RHS clearly has limit $0$, so by the pinching theorem (squeeze theorem, sandwich theorem)
$$\lim_{n\to\infty}a_n=0\ .$$
A: Let , $a_n=\frac{4^n}{(2n)!}$. Then  , $$\frac{a_{n+1}}{a_n}=\frac{4}{(2n+2)(2n+1)}\to 0 \text{ as } n\to \infty.$$
So, $$\lim_{n\to \infty}a_n=0.$$
A: Using AM GM inequality, $$\dfrac{2n}a\ge\dfrac{2n-r}a+\dfrac ra\ge2\sqrt{\dfrac{2n-r}a\cdot\dfrac ra}$$
$$\iff\dfrac{a^2}{(2n-r)r}\le\left(\dfrac an\right)^2$$
Set $r=1,2,\cdots, n$
