Answer
series converges when $|x|\lt 16$
and radius of convergence is $16$.
Work Step by Step
$\frac{1}{\sqrt[4] {16-x}}=\frac{1}{2}\Sigma_{n=1}^\infty\frac{1.5.9....(4n-3)x^{n}}{2^{6n+1}n!}$
$$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{1.5.9....(4n-3)x^{n+1}(4n-3+4)}{2^{6n+1}n!}}{\frac{1.5.9....(4n-3)x^{n}}{2^{6n+1}n!}}|$$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(4n-1)x}{2^{6}(n+1)}|$
$=\lim\limits_{n \to \infty}|\frac{4x-x/n}{2^{6}+2^{6}/n)}|$
$=|\frac{4x}{2^{6}}|$
$=|\frac{x}{2^{4}}|$
$=|\frac{x}{2^{4}}|\lt 1$
$=|x|\lt 16$
Thus, the series converges when $|x|\lt 16$
and radius of convergence is $16$.
