Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 4

$R=1$ ; interval of convergence is $(-1,1]$

Let $a_{n}=\frac {x^{n}}{\sqrt[3] {n}}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac {x^{n+1}}{\sqrt[3] {n+1}}}{\frac {x^{n}}{\sqrt[3] {n}}}|$ $=|x|$ converges if $|x|\lt 1$, so $R=1$ Hence, $R=1$ ; interval of convergence is $(-1,1]$