To determine the radius and interval of convergence for the given series

$\sum_{n=1}^{\infty} \frac{3x - 2}{n 3^n},$

we can apply the Ratio Test. The Ratio Test states that for a series $\sum a_n$, if

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L,$

then:

1. The series converges if $L < 1$,
2. The series diverges if $L > 1$,
3. The test is inconclusive if $L = 1$.

Step-by-Step Solution

1. Identify $a_n$: For the given series, the general term is: $a_n = \frac{3x - 2}{n 3^n}.$

2. Compute $\left| \frac{a_{n+1}}{a_n} \right|$: We need to find $a_{n+1}$ and then set up the ratio: $a_{n+1} = \frac{3x - 2}{(n+1) 3^{n+1}}.$ Now, we calculate the ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{3x - 2}{(n+1) 3^{n+1}}}{\frac{3x - 2}{n 3^n}} \right|.$

3. Simplify the Ratio: This simplifies to: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3x - 2}{3} \cdot \frac{n}{n+1} \right|.$ Taking the limit as $n \to \infty$, we get: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3x - 2}{3} \right|.$

4. Set up the inequality for convergence: For convergence, we require: $\left| \frac{3x - 2}{3} \right| < 1.$ Simplifying this inequality, we find: $-1 < \frac{3x - 2}{3} < 1.$ Multiply through by 3: $-3 < 3x - 2 < 3.$ Adding 2 to each part: $-1 < 3x < 5.$ Now, divide by 3: $-\frac{1}{3} < x < \frac{5}{3}.$

5. Conclusion: The radius of convergence is: $R = \frac{5}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}.$ The interval of convergence is: [