It's A Factorial Bonanza

Relevant wiki: Gamma Function

$\begin{aligned} S & = \sum_{k=0}^n \left(\frac {2k+1}2\right)! \left(-\frac {2k+3}2\right)! & \small \color{#3D99F6} \text{Gamma function }\Gamma(x+1) = x! \\ & = \sum_{k=0}^n \Gamma \left(\frac {2k+3}2\right) \Gamma \left(-\frac {2k+1}2\right) & \small \color{#3D99F6} \text{Note that } \Gamma(x)\Gamma(-x) = - \frac \pi{x\sin (\pi x)} \\ & = \sum_{k=0}^n - \frac {{\color{#3D99F6}\Gamma \left(\frac {2k+3}2\right)}\pi}{{\color{#3D99F6}\frac {2k+1}2\Gamma \left(\frac {2k+1}2\right)} \sin \left(\frac {2k+1}2\pi \right)} & \small \color{#3D99F6} \text{also } \Gamma(1+x) = z\Gamma(x) \\ & = \sum_{k=0}^n \frac {-\pi}{\sin \left(\frac {2k+1}2\pi \right)} & \small \color{#3D99F6} \text{Since } \sin \left(\frac {2k+1}2\pi\right) = \begin{cases} 1 & \text{if }x \text{ is even.} \\ -1 & \text{if }x \text{ is odd.} \end{cases} \\ & = \sum_{k=0}^n (-1)^{n+1}\pi = \begin{cases} -\pi & \text{if }n \text{ is even.} \\ \boxed 0 & \text{if }n \text{ is odd.} \end{cases} \end{aligned}$

Using the property that $r! = r(r-1)!$ for any $r \in \mathbb{R}\setminus\{0,-1,-2,-3,\ldots\},$ $\begin{aligned} \left(\frac{2(K+1)+1}{2}\right)!\left(\frac{-2(K+1)-3}{2}\right)! &= \left(\frac{2K+3}{2}\right)!\left(\frac{-2K-3}{2}-1\right)! \\ &= \left(\frac{2K+3}{2}\right)\left(\frac{2K+1}{2}\right)!\left(\frac{1}{\frac{-2K-3}{2}}\right)\left(\frac{-2K-3}{2}\right)! \\ &= (-1) \cdot \left(\frac{2K+1}{2}\right)!\left(\frac{-2K-3}{2}\right)! \end{aligned}$ which we can use as the inductive step to show for non-negative integral $K$ that $\left(\frac{2K+1}{2}\right)!\left(\frac{-2K-3}{2}\right)! = (-1)^K \left(\frac{1}{2}\right)!\left(-\frac{3}{2}\right)!$

Then $\sum_{K=0}^n \left(\frac{2K+1}{2}\right)!\left(\frac{-2K-3}{2}\right)! = \sum_{K=0}^n(-1)^K \left(\frac{1}{2}\right)!\left(-\frac{3}{2}\right)! = \begin{cases}0 & \text{ if } n \text{ is odd} \\ \left(\frac{1}{2}\right)!\left(-\frac{3}{2}\right)! & \text{ if } n \text{ is even}\end{cases}$

Since we're told $n$ is odd, the answer is $\boxed{0}$

Using the gamma function $\Gamma(p) = \int_{0}^{\infty} t^{p - 1} e^{-t} dt$ we obtain:

$\Gamma(\dfrac{1}{2}) = \displaystyle\int_{0}^{\infty} t^{-\frac{1}{2}} e^{-t} dt$

Let $s^2 = t \implies 2s ds = dt \implies$

$\Gamma(\dfrac{1}{2}) = 2\displaystyle\int_{0}^{\infty} e^{-s^2} ds$

Since s is a dummy variable we can write:

$(\Gamma(\dfrac{1}{2}))^2 =$ $4\displaystyle\int_{0}^{\infty} e^{-x^2} dx \int_{0}^{\infty} e^{-y^2} dy =$ $4\displaystyle\int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2 + y^2)} dx dy$

Let $x = r\cos\theta, y = r\sin\theta$

Using the Jacobian $\implies (\Gamma(\dfrac{1}{2}))^2 = 4\displaystyle\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-r^2} r dr d\theta =$ $-2\displaystyle\int_{0}^{\frac{\pi}{2}} e^{-r^2}|_{0}^{\infty} d\theta = 2\displaystyle\int_{0}^{\frac{\pi}{2}} d\theta = \pi \implies \Gamma(\dfrac{1}{2}) = \sqrt{\pi}$

Now, using integration by parts you can show that $\Gamma(p + 1) = p \Gamma(p)$

Show $\Gamma(1) = 1$ and recursively using $\Gamma(p + 1) = p \Gamma(p) \implies \Gamma(p + 1) = p!$ for any integer $p \geq 0$ and p can be extended to reals so that $\Gamma(\dfrac{1}{2}) = \Gamma(\dfrac{-1}{2} + 1) = \sqrt{\pi} = (\dfrac{-1}{2})!$

Using $\Gamma(p + 1) = p \Gamma(p)$ and $\Gamma(\dfrac{1}{2}) = \sqrt{\pi} = (\dfrac{-1}{2})!$

$\implies$

$S_{0} = \Gamma(\dfrac{3}{2}) = \dfrac{1}{2} * \Gamma(\dfrac{1}{2}) = \dfrac{\sqrt{\pi}}{2} = (\dfrac{1}{2})!$

$$

$S_{1} = \Gamma(\dfrac{5}{2}) = \dfrac{3}{2} * \Gamma(\dfrac{3}{2}) = \dfrac{1 * 3 * \sqrt{\pi}}{2^2} = (\dfrac{3}{2})!$

$S_{2} = \Gamma(\dfrac{7}{2}) = \dfrac{5}{2} * \Gamma(\dfrac{5}{2}) = \dfrac{1 * 3 * 5 * \sqrt{\pi}}{2^3} = (\dfrac{5}{2})!$

In General:

$S_{K} = \Gamma(\dfrac{2K + 3}{2}) = \dfrac{1 * 3 * 5 * * * (2K + 1) * \sqrt{\pi}}{2^{K + 1}} = (K + \frac{1}{2})!$ $$

$\implies (K + \dfrac{1}{2})! = \dfrac{(2K + 1)! * \sqrt{\pi}}{2^{2K + 1} * K!}$ , where $K$ is a non-negative integer. $$

Note: $(2K + 1)! = 1 * 2 * 3 * * * (2K) * (2K + 1) = 2^K * K! * (1 * 3 * 5 * * * (2K + 1)) \implies$ $1 * 3 * 5 * * * (2K + 1) = \dfrac{(2K + 1)!}{2^K * K!}$

and,

Using $\Gamma(p) = \dfrac{\Gamma(p + 1)}{p}$ and $\Gamma(\dfrac{1}{2}) = \sqrt{\pi} = (\dfrac{-1}{2})!$

$\implies$

$T_{0} = \Gamma(-\dfrac{1}{2}) = -2\Gamma(\dfrac{1}{2}) = -2\sqrt{\pi} = (-\dfrac{3}{2})!$

$T_{1} = \Gamma(-\dfrac{3}{2}) = -\dfrac{2}{3}\Gamma(-\dfrac{1}{2}) = \dfrac{2^2}{1 * 3}\sqrt{\pi} = (-\dfrac{5}{2})!$

$T_{2} = \Gamma(-\dfrac{5}{2}) = -\dfrac{2}{5}\Gamma(-\dfrac{3}{2}) = -\dfrac{2^3}{1 * 3 * 5}\sqrt{\pi} = (-\dfrac{7}{2})!$

In General:

$T_{K} = \Gamma(-K -\dfrac{1}{2}) = \pm\dfrac{2^{K + 1}\sqrt{\pi}}{1 * 3 * 5 * * * (2K + 1)} = (-K -\dfrac{3}{2})!$ $$

$\implies (-K - \dfrac{3}{2})! = \pm\dfrac{2^{2K + 1} * K!\sqrt{\pi}}{(2K + 1)!}$ , where $K$ is a non-negative integer

$\implies$

$S_{K} * T_{K} =$ $\begin{cases} \pi & \text{for} \:\ K \:\ \text{odd}\\ -\pi & \text{for} \:\ K \:\ \text{even}\\ \end{cases}$

$\therefore n$ odd $\implies$ $\displaystyle\sum_{K = 0}^{n} \left(\left(\frac{2K + 1}{2}\right)! \left(\frac{-2K - 3}{2}\right)!\right) = \displaystyle\sum_{K = 0}^{n} S_{K} * T_{K} = \boxed{0}$ .

$$

$$

Note: It follows that $n$ even $\implies \displaystyle\sum_{K = 0}^{n} \left(\left(\frac{2K + 1}{2}\right)! \left(\frac{-2K - 3}{2}\right)!\right) = -\pi$ .