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The problem you have uploaded asks to find the interval of convergence for the series:

\[
\sum_{n=0}^{\infty} \frac{(2x)^n}{3n+1}
\]

To determine the interval of convergence of a power series, we generally use the **Ratio Test** or **Root Test** to find the radius of convergence. However, in this case, since the general term has a more complex denominator \((3n + 1)\), we will need to carefully apply the **Ratio Test** and analyze the behavior of the terms.

### Step-by-Step Solution:

1. **Identify the general term** of the series:

   \[
   a_n = \frac{(2x)^n}{3n + 1}
   \]

2. **Apply the Ratio Test**, which is given by:

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

   Substituting the general terms \(a_{n+1}\) and \(a_n\), we get:

   \[
   \frac{a_{n+1}}{a_n} = \frac{\frac{(2x)^{n+1}}{3(n+1) + 1}}{\frac{(2x)^n}{3n + 1}} = \frac{(2x)^{n+1}}{(2x)^n} \cdot \frac{3n + 1}{3n + 4}
   \]

   Simplifying this expression:

   \[
   \frac{a_{n+1}}{a_n} = 2x \cdot \frac{3n + 1}{3n + 4}
   \]

3. **Take the limit as \(n \to \infty\)**:

   \[
   \lim_{n \to \infty} \left| 2x \cdot \frac{3n + 1}{3n + 4} \right| = \lim_{n \to \infty} \left| 2x \cdot \frac{3n}{3n} \right| = \left| 2x \right|
   \]

4. **Apply the Ratio Test's convergence condition**:

   The Ratio Test states that the series converges when:

   \[
   \left| 2x \right| < 1 \quad \Rightarrow \quad |x| < \frac{1}{2}
   \]

   This gives the radius of convergence \(R = \frac{1}{2}\).

5. **Determine the interval of convergence**:

   The series converges when \(|x| < \frac{1}{2}\), so the interval of convergence is initially:

   \[
   -\frac{1}{2} < x < \frac{1}{2}
   \]

6. **Check the endpoints \(x = -\frac{1}{2}\) and \(x = \frac{1}{2}\)** by substitution into the original series**:

   - For \(x = \frac{1}{2}\), the series becomes:

     \[
     \sum_{n=0}^{\infty} \frac{(2 \cdot \frac{1}{2})^n}{3n + 1} = \sum_{n=0}^{\infty} \frac{1^n}{3n + 1} = \sum_{n=0}^{\infty} \frac{1}{3n + 1}
     \]

     This series does **not converge**, as it behaves like a harmonic series for large \(n\).

   - For \(x = -\frac{1}{2}\), the series becomes:

     \[
     \sum_{n=0}^{\infty} \frac{(2 \cdot -\frac{1}{2})^n}{3n + 1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{3n + 1}
     \]

     This is an **alternating series**, and we need to apply the **Alternating Series Test**. For large \(n\), the terms \(\frac{1}{3n+1}\) tend to zero, and the series alternates. Hence, the series **converges** at \(x = -\frac{1}{2}\).

### Conclusion:

The interval of convergence is:

\[
\boxed{[-\frac{1}{2}, \frac{1}{2})}
\]

Would you like more details on any of the steps or have further questions?

---

Here are some related questions to further your understanding:

1. How does the Ratio Test help in finding the radius of convergence?
2. Why do we need to check the endpoints separately in convergence problems?
3. How does the Alternating Series Test work, and why does it apply to this problem at \(x = -\frac{1}{2}\)?
4. What is the difference between absolute and conditional convergence in series?
5. Can you explain how harmonic series behave and why they fail to converge?

**Tip**: When using the Ratio Test, remember that the series converges absolutely if the limit is less than 1 and diverges if the limit is greater than 1. Always check the boundary points separately!

Find the interval of convergence for the sum: ∑(2x)^n / (3n + 1) from n = 0 to infinity.

Learn how to find the interval of convergence for the series ∑(2x)^n / (3n + 1) using the Ratio Test. This problem covers key series concepts and tests for convergence. Suitable for high school or early undergraduate students.

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