Answer
$\displaystyle \frac{1}{2}$
Work Step by Step
$\displaystyle \sum_{n=0}^{\infty}(\frac{1}{2^{n}}-\frac{1}{3^{n}})=\sum_{n=0}^{\infty}(\frac{1}{2})^{n}-\sum_{n=0}^{\infty}(\frac{1}{3})^{n}$
Both are geometric series,
the first with $a=1, r=\displaystyle \frac{1}{2} < 1$
the second with $a=1, r=\displaystyle \frac{1}{3} < 1$.
(both converge by Th 9/6, $\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$)
$\displaystyle \sum_{n=0}^{\infty}(\frac{1}{2^{n}}-\frac{1}{3^{n}})=\frac{1}{1-(\frac{1}{2})}-\frac{1}{1-(\frac{1}{3})}$
$=\displaystyle \frac{1}{\frac{1}{2}}-\frac{1}{\frac{2}{3}}$
$=2-\displaystyle \frac{3}{2}$
$=\displaystyle \frac{1}{2}$
