Factorial Fest

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

$\implies$

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

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

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

In General:

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

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

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

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

For $K \in \mathbb{N}$

$\displaystyle\sum_{J = 2}^{K + 1} ( (-K - \dfrac{3}{2})! + J ) = \dfrac{(K + 2)(K + 1)}{2} - 1 + K * (-K - \dfrac{3}{2})! = \dfrac{K(K + 3)}{2} \pm \dfrac{2^{2K + 1} * K * K!\sqrt{\pi}}{(2K + 1)!}$

$\therefore \displaystyle\sum_{J = 2}^{1000} ((-999 - \dfrac{3}{2})! + J) = \displaystyle\sum_{J = 2}^{1000} \left(\left(\frac{-2001}{2}\right)! + J\right) =$

$999 * 501 + \dfrac{999 * (999)! * 2^{1999}\sqrt{\pi}}{(1999)!} = 500499 + \dfrac{999 * (999)! * 2^{1999}\sqrt{\pi}}{(1999)!} = a + \dfrac{b * b! * 2^c\sqrt{\pi}}{c!} \implies a + b + c = \boxed{503497}$ .