I Was Very Amazed At The Solution 9

I used the theorem that for any $\alpha$ ,

$\sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} = \dfrac{1}{\alpha} \cdot$

Hence, the equation above can be simplified to:

$16\left( \sum_{\alpha \ \in \ \text{{positive factors of 360}}} \dfrac{1}{\alpha} \right ) \cdot$

But, as $\displaystyle \sum_{\alpha \ \in \ \text{{positive factors of 360}}} \dfrac{1}{\alpha} = \dfrac{\text{sum of positive factors of 360}}{360} = \dfrac{1170}{360} \ ,$

$\begin{aligned} \implies 16 \left( \sum_{\alpha \ \in \ \text{{positive factors of 360}}} \left(\sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} \right)\right) \\ &= 16\left(\dfrac{1170}{360}\right) \\ &= \boxed{52} \end{aligned}$