smc

The infinite sum of a geometric series with first term $a$ and common ratio $r$ ($-1<r<1$) is $\frac{a}{1-r}$. Now, in this geometric series, $a=4$, and $r=-\frac{2}{3}$. Plugging these into the formula, we get $\frac{4}{1-(-\frac{2}{3})}$, which simplifies to $\frac{12}{5}$, or $\boxed{2.4}$.