I can do part a) and b) but I'm not getting how to do part c)

So I'm guessing the part under the square root cannot be negative, and so $\displaystyle \dfrac{64b^2}{a^4}x^2 \le 1$

Solving that I would get:

$\displaystyle x^2 \le \dfrac{64b^2}{a^4}$

$\displaystyle |x| \le \dfrac{8b}{a^2}$

And so the radius of convergence is: $\displaystyle \dfrac{8b}{a^2}$ ?

And how do I show that for the series to converge, $\displaystyle b < \dfrac{a}{4}$ ?