Answer
series converges when $|x|\lt \frac{1}{3}$
and radius of convergence is $\frac{1}{3}$.
Work Step by Step
$f(x)=(1-3x)^{-5}=(1+(-3x))^{-5}=1+\Sigma_{n=1}^\infty\frac{5.6.7....(n+4)3^{n}x^{n}}{n!}$
$$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{5.6.7....(n+4)(n+5)3^{n+1}x^{n+1}}{n+1!}}{\frac{5.6.7....(n+4)3^{n}x^{n}}{n!}}|$$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(n+5)3x}{n+1}|$
$=\lim\limits_{n \to \infty}|\frac{3x+15x/n}{1+1/n}|$
$=|3x|$
$=|3x|\lt 1$
$=|x|\lt \frac{1}{3}$
Thus, the series converges when $|x|\lt \frac{1}{3}$
and radius of convergence is $\frac{1}{3}$.
