1. **State the problem:** We are given the series $$S = \sum_{n=1}^\infty \frac{1 \cdot 4 \cdot 7 \cdots (3n - 2)}{2 \cdot 5 \cdot 8 \cdots (3n - 1)}$$ and need to determine whether it converges or diverges. 2. **Rewrite the general term:** The numerator is the product of terms of the form $(3k - 2)$ for $k=1$ to $n$, and the denominator is the product of terms $(3k - 1)$ for $k=1$ to $n$. So the $n$th term is: $$a_n = \frac{\prod_{k=1}^n (3k - 2)}{\prod_{k=1}^n (3k - 1)}$$ 3. **Analyze the ratio of terms:** To understand the behavior, consider the ratio of consecutive terms: $$\frac{a_{n+1}}{a_n} = \frac{3(n+1) - 2}{3(n+1) - 1} = \frac{3n + 1}{3n + 2}$$ 4. **Limit of the ratio:** As $n \to \infty$, $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{3n + 1}{3n + 2} = 1$$ 5. **Apply the ratio test:** Since the limit of the ratio is 1, the ratio test is inconclusive. 6. **Examine the term $a_n$ itself:** For large $n$, approximate each term: $$a_n = \prod_{k=1}^n \frac{3k - 2}{3k - 1} = \prod_{k=1}^n \left(1 - \frac{1}{3k - 1}\right)$$ 7. **Use logarithms to analyze the product:** $$\ln a_n = \sum_{k=1}^n \ln\left(1 - \frac{1}{3k - 1}\right)$$ For large $k$, use $\ln(1 - x) \approx -x$: $$\ln a_n \approx - \sum_{k=1}^n \frac{1}{3k - 1}$$ 8. **Behavior of the sum:** The sum $\sum_{k=1}^n \frac{1}{3k - 1}$ behaves like a constant times the harmonic series: $$\sum_{k=1}^n \frac{1}{3k - 1} \sim \frac{1}{3} \sum_{k=1}^n \frac{1}{k}$$ which diverges to infinity as $n \to \infty$. 9. **Conclusion on $a_n$:** Since $$\ln a_n \approx - \frac{1}{3} \ln n + \text{constant}$$ we have $$a_n \approx \frac{C}{n^{1/3}}$$ for some constant $C > 0$. 10. **Test convergence of the series:** The series behaves like $$\sum_{n=1}^\infty \frac{1}{n^{1/3}}$$ which diverges because the $p$-series converges only if $p > 1$. **Final answer:** The series $S$ is **divergent**.