Answer
$R=\frac{1}{4}$ ; the interval of convergence is $(-\frac{1}{4},\frac{1}{4}]$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{4^{n+1}x^{n+1}}{\sqrt {(n+1)}}}{\frac{4^nx^n}{\sqrt n}}|$
$=4|x|$
Hence, $R=\frac{1}{4}$ ; the interval of convergence is $(-\frac{1}{4},\frac{1}{4}]$
