## Calculus: Early Transcendentals 8th Edition

The series converges if $-1 \lt x \lt 5$ and the sum is $\frac{3}{5-x}$
$\Sigma^{\infty}_{n=0} \frac{(x-2)^{n}}{3^{n}}$ $\Sigma^{\infty}_{n=0} (\frac{x-2}{3})^{n}$ $a=1$ and $r=(\frac{x-2}{3})$ Therefore, the series is convergent. $|r| \lt 1$ $|\frac{x-2}{3}| \lt 1$ $|x-2| \lt 3$ $-3 \lt x-2 \lt 3$ $-1 \lt x \lt 5$ The sum $\Sigma^{\infty}_{n=1} \frac{(x-2)^{n}}{3^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-(\frac{x-2}{3})}$ $=\frac{1}{(\frac{3-x+2}{3})}$ $=\frac{3}{5-x}$