Answer
$\dfrac{a}{1-r}$
Work Step by Step
If $|r|<1$, the sum of the geometric series $\sum_{k=1}^{\infty} ar^{k-1}$ is $\dfrac{a}{1-r}$ because for the geometric series $ar^{k-1}$ with first term $a_1=ar^0=a$ and ratio $r$ where $|r|<1$, the sum is $\sum_{k=1}^{\infty} a_1r^{k-1}=\dfrac{a_1}{1-r}$.