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!
 A: 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 $n\rightarrow {\infty}$.
A: After some pondering i'm going to try to answer this myself.
I've been thinking about considering two cases for $n$ is odd and $n$ is even. Start with even $n$. We know that for any $n \ge 2 \implies C_{3/2}^n > 0$ in case $n$ is even. Lets inspect a subsequence for even indices:
$$
C_{3/2}^n = \frac{1}{2^nn!}\prod_{k=1}^n(5-2k) \\
C_{3/2}^{n+2} = \frac{1}{2^{n+2}(n+2)!}\prod_{k=1}^{n+2}(5-2k)
$$
Consider the fraction:
$$
\frac{C_{3/2}^{n+2}}{C_{3/2}^n} = \frac{2^nn!}{2^{n+2}(n+2)!}(5-2(n+1))(5-2(n+2)) = \\
=\frac{1}{2^2(n+1)(n+2)}\cdot(3-2n)(1-2n) = \frac{(2n-3)(2n-1)}{(2n+2)(2n+4)}
$$
Clearly this is less than $1$ and hence the sequence is decreasing towards $0^+$. With that being said the maximum is obtained for $n=1$ which is $3 \over 2$. Also note that for odd indices the value for $C_{3/2}^n < 0$.
Now consider the case for odd $n$:
$$
C_{3/2}^n = -\left|\frac{1}{2^nn!}\prod_{k=1}^n(5-2k)\right| \\
C_{3/2}^{n+2} = -\left|\frac{1}{2^{n+2}(n+2)!}\prod_{k=1}^{n+2}(5-2k)\right|
$$
Now using the results from the case for even $n$ we obtain that:
$$
\frac{|C_{3/2}^{n+2}|}{|C_{3/2}^{n}|} < 1
$$
At the same time $C_{3/2}^n$ for odd $n$ is less than $0$ hence for odd $n$:
$$
0 < C_{3/2}^{n} < C_{3/2}^{n+2}
$$
This means $C_{3/2}^n$ is increasing and tending to $0^-$. Thus the minimum value of $C_{3/2}^n$ is obtained at $n=3$
Compiling both cases gives us that:
$$
C_{3/2}^3< C_{3/2}^n < C_{3/2}^1
$$
or:
$$
−0.0625 < C_{3/2}^n < {3 \over 2}
$$
I would appreciate any comments on the given approach.
