# Can the following sum involving double factorial be simplified?

I have the following sum: $$\sum_{k=2}^{n-1}\frac{(n-k)(2k-3)!!}{a^k}$$ where $p!!$ is the double factorial defined as $$p!!=\left\{\begin{array}{} p\cdot (p-2)\cdots 3\cdot 1, & \mbox{if p is odd}\\ p\cdot (p-2)\cdots 4\cdot 2, & \mbox{if p is even} \end{array}\right.$$

It can see that it can be expressed (up to some multiplication factor) as a convolution of the two finite sequences $\{ka^k\}_{k=2}^{n-1}$ and $\{(2k-3)!!\}_{k=2}^{n-1}$, but I am not sure what I can do next to simplify this. Please help.