1. The problem is to find the sum of the series: $\frac{1}{6} + \frac{1}{2} + \frac{3}{2} + \ldots$ 2. First, identify the pattern or rule for the terms in the series. 3. The terms are $\frac{1}{6}, \frac{1}{2}, \frac{3}{2}, \ldots$ 4. Let's write the terms with a common denominator to see the pattern better: $$\frac{1}{6}, \frac{3}{6}, \frac{9}{6}, \ldots$$ 5. Notice the numerators form a geometric sequence: $1, 3, 9, \ldots$ with common ratio $3$. 6. The series can be expressed as: $$\sum_{n=0}^\infty \frac{3^n}{6} = \frac{1}{6} \sum_{n=0}^\infty 3^n$$ 7. However, the sum $\sum_{n=0}^\infty 3^n$ diverges because the common ratio $3 > 1$. 8. Therefore, the series $\frac{1}{6} + \frac{1}{2} + \frac{3}{2} + \ldots$ diverges and does not have a finite sum. Final answer: The series diverges and the sum is infinite.