Answer
$8$
Work Step by Step
An infinite geometric series has a sum if and only if $|r|\lt1$, where $r$ is the common ratio. If it exists, then it equals $\frac{a_1}{1-r}$ where $a_1$ is the first term.
Here $r=\frac{5}{8},a_1=3$
Thus $|\frac{5}{8}|\lt1$, hence the sum exists.
Hence the sum: $\dfrac{3}{1-\frac{5}{8}}=8$
