Answer:
The geometric series that diverges is the one with a common ratio greater than or equal to 1 in absolute value. Let's check each series:
1. Three-fifths + three-tenths + three-twentieths + ...
Common ratio: \( \frac{3/10}{3/5} = \frac{1}{2} \)
Since \( \frac{1}{2} < 1 \), this series converges.
2. Negative 10 + 4 - eight-fifths + \( \frac{16}{25} \) - ...
Common ratio: \( \frac{4}{-10} = -\frac{2}{5} \)
Since \( \left| -\frac{2}{5} \right| < 1 \), this series converges.
3. \( \sum_{n=1}^{\infty} \frac{2}{3}(-4)^{n-1} \)
Common ratio: \( -\frac{4}{3} \)
Since \( \left| -\frac{4}{3} \right| > 1 \), this series diverges.
4. \( \sum_{n=1}^{\infty} (-12)\left(\frac{1}{5}\right)^{n-1} \)
Common ratio: \( \frac{1}{5} \)
Since \( \frac{1}{5} < 1 \), this series converges.