Define the double factorial of $n$, denoted $n!!$, as follows: $$n!! = \begin{cases} 1 \cdot 3 \cdot 5 \cdot \cdot \cdot \cdot (n-2) \cdot n \quad \text{if } n \text{ is odd } \\ 2 \cdot 4 \cdot 6 \cdot \cdot \cdot \cdot (n-2) \cdot n \quad \text{if } n \text{ is even } \end{cases}$$ and $(-1)!! = 0!! = 1$.

Find the radius of convergence for the given power series. $$\sum_{n=1}^\infty \frac { 3^n \cdot n! \cdot (2 n + 8)! \cdot (2n)!! } { 6^n \cdot [(n + 9)!]^2 \cdot (4n - 3)!!} (-7 x + 2)^n$$
The radius of convergence, $R =$