## Calculus 8th Edition

$R=\frac{1}{4}$ and Interval of convergence: $(-\frac{1}{4},\frac{1}{4}]$
Given: $a_n=\frac{4^nx^n}{\sqrt n}$ Thus, $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{4^{n+1}x^{n+1}}{\sqrt {(n+1)}}}{\frac{4^nx^n}{\sqrt n}}|=4|x|$ Hence, $R=\frac{1}{4}$ and Interval of convergence: $(-\frac{1}{4},\frac{1}{4}]$