Answer
$R=\frac{1}{5}$ ; interval of convergence is $[\frac{3}{5},1)$
Work Step by Step
Let $a_{n}=\frac{(5x-4)^{n}}{n^{3}}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(5x-4)^{n+1}}{(n+1)^{3}}}{\frac{(5x-4)^{n}}{n^{3}}}|$
$=|5x-4|$
$=|5x-4|\lt 1$
Hence, $R=\frac{1}{5}$ ; interval of convergence is $[\frac{3}{5},1)$
