Here is the basic idea: $\left( {\forall \left| r \right| < 1} \right)\left[ {\sum\limits_{k = 0}^\infty {a{r^k}} = \frac{a}{{1 - r}}} \right]$
Notice that $\sum\limits_{k = 1}^\infty {4{{\left[ {\frac{1}{3}} \right]}^{k - 1}}} = \sum\limits_{k = 0}^\infty {\frac{4}{{{3^k}}}}$