Answer
The series is convergent and its sum is $\frac{32}{7}$.
Work Step by Step
$\Sigma^{\infty}_{n=1}[(0.8)^{n-1} - (0.3)^{n}]$
For $\Sigma^{\infty}_{n=1}(0.8)^{n-1}$, $a=1$ and $r=-0.8$
Since $|r|= 0.8 \lt 1$, therefore $\Sigma^{\infty}_{n=1}(0.8)^{n-1}$ is convergent and its sum is
$=\frac{a}{1-r}$
$=\frac{1}{1-0.8}$
$=\frac{1}{0.2}$
$=5$
For $\Sigma^{\infty}_{n=1}(0.3)^{n} = \Sigma^{\infty}_{n=1}(0.3)(0.3)^{n-1}$, $a=0.3$ and $r=0.3$
Since $|r|=0.3 \lt 1$, the series $\Sigma^{\infty}_{n=1}(0.3)^{n}$ converges and its sum is
$=\frac{a}{1-r}$
$=\frac{0.3}{1-0.3}$
$=\frac{3}{7}$
Since both sums are convergent, the series is convergent and its sum is $5-\frac{3}{7} = \frac{32}{7}$
