Answer
The series converges to $\frac{448}{3}$.
Work Step by Step
Step 1
This is a geometric series with a common ratio $r = \frac{4}{7}$. Since $|r| < 1$, this series converges.
The initial term is
$a = \frac{4^3}{7^0} = \frac{64}{1} = 64$.
Step 2
The sum is:
$S= \frac{a_1}{1 - r} = \frac{64}{1 - \frac{4}{7}} = 64\cdot \frac{7}{3}= \frac{448}{3}$.
Result The series converges to $\frac{448}{3}$.
