## Calculus 8th Edition

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