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To find the interval of convergence for the series

\[
\sum_{n=0}^{\infty} \frac{(3x - 1)^n}{n^3 + n},
\]

we can use the **Ratio Test** to determine where the series converges. The Ratio Test states that for a series \(\sum a_n\), the series converges absolutely if

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.
\]

### Step-by-Step Solution

1. **Identify \(a_n\):** In this case,
   \[
   a_n = \frac{(3x - 1)^n}{n^3 + n}.
   \]

2. **Apply the Ratio Test:**
   We need to compute
   \[
   \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.
   \]
   
   Substituting \(a_n\) and \(a_{n+1}\), we have
   \[
   \frac{a_{n+1}}{a_n} = \frac{\frac{(3x - 1)^{n+1}}{(n+1)^3 + (n+1)}}{\frac{(3x - 1)^n}{n^3 + n}} = \left| (3x - 1) \right| \cdot \frac{n^3 + n}{(n+1)^3 + (n+1)}.
   \]

3. **Simplify the Expression:**
   Now, let's analyze the limit:
   \[
   \lim_{n \to \infty} \left| (3x - 1) \right| \cdot \frac{n^3 + n}{(n+1)^3 + (n+1)}.
   \]

   As \(n \to \infty\), the highest power terms in both the numerator and denominator dominate, so we approximate:
   \[
   \frac{n^3 + n}{(n+1)^3 + (n+1)} \approx \frac{n^3}{n^3} = 1.
   \]

   Thus, the limit simplifies to
   \[
   \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| 3x - 1 \right|.
   \]

4. **Set Up the Convergence Condition:**
   For convergence, the Ratio Test requires
   \[
   \left| 3x - 1 \right| < 1.
   \]

5. **Solve for \(x\):**
   Breaking down the inequality:
   \[
   -1 < 3x - 1 < 1.
   \]
   Adding 1 to each part:
   \[
   0 < 3x < 2.
   \]
   Dividing by 3:
   \[
   0 < x < \frac{2}{3}.
   \]

### Conclusion
The interval of convergence for the series is

\[
\left( 0, \frac{2}{3} \right).
\]

Would you like further details on the steps, or do you have any questions?

---

Here are some related questions for deeper exploration:

1. How does the Ratio Test compare to other convergence tests, such as the Root Test?
2. Why is it necessary to check endpoints separately when determining interval convergence?
3. Could you explain why the Ratio Test works particularly well with power series?
4. What would the interval of convergence be if we had a different coefficient for \(x\) inside the series?
5. How can you determine if a series is absolutely or conditionally convergent?

**Tip:** When using the Ratio Test, remember to simplify terms carefully at each step, especially when dividing by expressions involving \(n\).

Find the intervals of convergence. Provide your final answer in interval notation. ∑ (3x - 1)^n / (n^3 + n)

Discover how to find the interval of convergence for the power series ∑ (3x - 1)^n / (n^3 + n) using the Ratio Test. This problem provides a detailed solution to determine where the series converges, focusing on power series and convergence concepts. Suitable for college-level mathematics.

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