The series is:
$\Sigma_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} (x+8)^n$
My attempt: let $a_n = \frac{(n!)^2}{(2n)!}$. Then $\rho = \lim_{n\rightarrow \infty} \vert \frac{a_{n+1}}{a_n} \vert = \frac{1}{4}$. Then the radius of convergence is $(-12,-4)$.
I got stuck at concluding the convergence at the end-points.
At $x = -4, \Sigma_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} (4)^n$. I tried the d'Alembert and the limit is $1$ so I couldn't conclude.
At $x = -12, \Sigma_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} (-4)^n$, which is an alternating number series. I only learned the Leibniz theorem and couldn't see how to apply it in this case.
The radius of convergence is a number (or $+\infty$), not an interval. In this case, this number is $4$.
It turns out that $(\forall n\in\mathbb{N}):4^na_n\geqslant1$. Therefore, your series diverges it both extremes, since we don't have $\lim_{n\to\infty}4^na_n=0$.