Prove that $C_{3 \over 2}^n$ is bounded given $C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}$ 
Let: 
  $$
\begin{cases}
C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}\\
C_{a}^0 = 1
\end{cases}
$$
  Prove $C_{3 \over 2}^n$ is bounded.

I've started with finding a reduced formula:
$$
C_{3\over 2}^n = \frac{{3\over 2}\left({3\over 2} - 1\right)\left({3\over 2} - 2\right)\left({3\over 2} - 3\right)\cdots\left({3\over 2} - n+1\right)}{n!} =\\
= \frac{3\cdot 1 \cdot(-1)\cdot(-3)\cdots(3-2n)}{2^n\cdot n!} = \\
\frac{1}{2^n \cdot n!}\prod_{k=1}^n\left(5-2k\right)
$$
This seems to converge to $0$ and therefore the sequence should be bounded, but how do i formally show that using inequalities? Moreover the elements change their sign depending on $n$ is even/odd (should i consider absolute values?).
Please note the precalculus tag. Thank you!

We should start from using the known expression of Gamma function: 
$\Gamma(z)= \frac{\Gamma(z+n+1)}{z(z+1)....(z+n)}$  following that $z(z+1)....(z+n-1) =\frac{\Gamma(z+n+1)}{\Gamma(z) (n+z)}$
On the other hand: 
$C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}=(-1)^n\frac{(-a)(-a+1)(-a+2)\dots(-a+n-1)}{n!}$
If $z=-a=-\frac{3}{2}$ put it back to $\Gamma(z)$ we get:
$C_{\frac{3}{2}}^n =\frac{(-1)^n}{n!}\frac{\Gamma(n-\frac{1}{2})}{\Gamma(-\frac{3}{2}) (n-\frac{3}{2})}$ 
As $\Gamma(n-\frac{1}{2})=\frac{\Gamma(n+\frac{1}{2})}{(n-\frac{1}{2})}$ using that $\Gamma(n+\frac{1}{2})=\frac{(2n)! \sqrt{\pi}}{4^n n!}$ and $\Gamma(-\frac{3}{2})=\frac{4\sqrt{\pi}}{3}$ 
We have the following: 
$C_{\frac{3}{2}}^n =3\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{1}{2n-1}\frac{1}{2n-3}$
We handle the even and odd terms separately and perform the ratio test for both terms: 
$n={2m}$, if n - even and ${2m+1}$, if n - odd. 
Let's take $\frac{C^{2m+2}_{\frac{3}{2}}}{C^{2m}_{\frac{3}{2}}}$ and $\frac{C^{2m+3}_{\frac{3}{2}}}{C^{2m+1}_{\frac{3}{2}}}$
After perform the simplifications we have: 
$\frac{C^{2m+2}_{\frac{3}{2}}}{C^{2m}_{\frac{3}{2}}}=\frac{(4m-1)(4m-3)}{(4m+2)(4m+4)}\lt 1$ so the even terms are decreasing and tends to zero, because $C_{\frac{3}{2}}^{2m} \gt 0$ 
$\frac{C^{2m+3}_{\frac{3}{2}}}{C^{2m+1}_{\frac{3}{2}}}=\frac{(4m+1)(4m-1)}{(4m+6)(4m+4)}\lt 1$ so the odd terms are increasing and tends to zero, because $C_{\frac{3}{2}}^{2m+1} \lt 0$
So $C_{\frac{3}{2}}^n \rightarrow{0}$ if $n\rightarrow {\infty}$.
