## Calculus 8th Edition

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