Answer
$-\Sigma_{n=0}^{\infty}\frac{(x+3)^{n}}{3^{n+1}}$, $R=3$
Work Step by Step
$\frac{1}{x}=-\Sigma_{n=0}^{\infty}\frac{(x+3)^{n}}{3^{n+1}}$
$\lim\limits_{n \to \infty}|\frac{\frac{(x+3)^{n+1}}{3^{n+2}}}{\frac{(x+3)^{n}}{3^{n+1}}}|$
$=\lim\limits_{n \to\infty}|\frac{x+3}{3}|\lt 1$
$|x+3|\lt 3$
$-3\lt x+3 \lt 3$
$-6 \lt x\lt 0$
The radius of convergence is always half of the width of the interval, so $R=3$.
