Answer
$R=1$ ; interval of convergence is $(-1,1]$
Work Step by Step
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 $(-1,1]$