Answer
$\frac{7}{3}$
Work Step by Step
We can split the initial series into the sum of two separate geometric series and evaluate their sums separately: $\sum^{\infty}_{n=1} (-0.2)^n$ + $\sum^{\infty}_{n=1} (0.6)^{n-1}$
We can see that $\sum^{\infty}_{n=1} (-0.2)^n$ is a geometric series, with initial term $ a = 0.2$ and $r = -0.2$
We use the geometric series formula $$\sum^{\infty}_{n=0} ar^n = \frac{a}{1-r}$$ which gives us $\frac{-0.2}{1-(-0.2)} = \frac{-0.2}{1.2}$
Likewise, we can see that $\sum^{\infty}_{n=1} (0.6)^{n-1}$ is a geometric series, with initial term $ a = 1$ and $r = 0.6$
Using the same formula, we get $\frac{1}{1-0.6} = \frac{1}{0.4}$
We now add $ \frac{-0.2}{1.2}$ and $\frac{1}{0.4}$ together and get $\frac{-0.2 + 3}{1.2} = \frac{2.8}{1.2} = \frac{7}{3} $
