To find the interval of convergence for the series $\sum_{n=1}^\infty \frac{(3x-1)^n}{n^3 + n}$, we can apply the Ratio Test to determine the radius of convergence. Let's work through this step by step.

Step 1: Apply the Ratio Test

The general term of the series is:

$a_n = \frac{(3x - 1)^n}{n^3 + n}.$

The Ratio Test tells us to examine the limit of the ratio of consecutive terms as $n \to \infty$:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(3x - 1)^{n+1}}{(n+1)^3 + (n+1)} \cdot \frac{n^3 + n}{(3x - 1)^n} \right|.$

Simplifying the expression:

$L = \lim_{n \to \infty} \left| (3x - 1) \cdot \frac{n^3 + n}{(n+1)^3 + (n+1)} \right|.$

Now, we will focus on simplifying the second term:

$\frac{n^3 + n}{(n+1)^3 + (n+1)} = \frac{n^3 + n}{n^3 + 3n^2 + 3n + 1 + n + 1} = \frac{n^3 + n}{n^3 + 3n^2 + 4n + 2}.$

As $n \to \infty$, this fraction approaches 1 because the highest powers of $n$ in both the numerator and denominator are the same (both have $n^3$). Therefore, the limit simplifies to:

$L = |3x - 1| \cdot 1 = |3x - 1|.$

Step 2: Set up the convergence condition

For the series to converge, the Ratio Test tells us that $L < 1$. Hence, we need:

$|3x - 1| < 1.$

Step 3: Solve the inequality

Solving $|3x - 1| < 1$:

$-1 < 3x - 1 < 1.$

Add 1 to all parts of the inequality:

$0 < 3x < 2.$

Now, divide by 3:

$0 < x < \frac{2}{3}.$

So, the series converges when $x$ is in the interval $(0, \frac{2}{3})$.

Step 4: Test the endpoints

We need to check the behavior of the series at the endpoints $x = 0$ and $x = \frac{2}{3}$.

At $x = 0$:

Substitute $x = 0$ into the series:

$\sum_{n=1}^\infty \frac{(3(0) - 1)^n}{n^3 + n} = \sum_{n=1}^\infty \frac{(-1)^n}{n^3 + n}.$

This is an alternating series. The general term tends to 0 as $n \to \infty$, and the terms decrease in magnitude, so by the Alternating Series Test, the series converges at $x = 0$.

At $x = \frac{2}{3}$:

Substitute $x = \frac{2}{3}$ into the series:

$\sum_{n=1}^\infty \frac{(3(\frac{2}{3}) - 1)^n}{n^3 + n} = \sum_{n=1}^\infty \frac{0^n}{n^3 + n} = 0.$

Since the entire series is 0, the series trivially converges at $x = \frac{2}{3}$.

Step 5: Conclusion

The series converges for $x$ in the interval $[0, \frac{2}{3}]$. Thus, the interval of convergence is:

$\boxed{[0, \frac{2}{3}]}.$

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Would you like further details on any step, or do you have any other questions? Here's a list of related topics you might find interesting:

1. How to apply the Root Test for series convergence.
2. The Alternating Series Test in more detail.
3. How to use the Ratio Test for series with variable exponents.
4. Understanding the concept of radius and interval of convergence.
5. The behavior of power series at the endpoints of their interval of convergence.

Tip: When working with series, always check the endpoints separately, even if the ratio test gives a strict inequality for convergence.