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To determine whether the series 

\[
\sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1}
\]

is absolutely convergent or divergent using the ratio test, we proceed as follows:

### Step-by-Step Solution

1. **Identify the General Term**:
   The general term of the series is
   \[
   a_n = \frac{n(x-4)^n}{n^3 + 1}.
   \]

2. **Set Up the Ratio Test**:
   The ratio test states that for a series \(\sum a_n\), we look at
   \[
   L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.
   \]
   - If \(L < 1\), the series converges absolutely.
   - If \(L > 1\), the series diverges.
   - If \(L = 1\), the test is inconclusive.

3. **Calculate \(\left| \frac{a_{n+1}}{a_n} \right|\)**:
   First, find \(a_{n+1}\):
   \[
   a_{n+1} = \frac{(n+1)(x-4)^{n+1}}{(n+1)^3 + 1}.
   \]
   
   Now compute the ratio \(\frac{a_{n+1}}{a_n}\):
   \[
   \frac{a_{n+1}}{a_n} = \frac{(n+1)(x-4)^{n+1}}{(n+1)^3 + 1} \cdot \frac{n^3 + 1}{n(x-4)^n}.
   \]
   Simplifying this expression:
   \[
   \frac{a_{n+1}}{a_n} = \left| \frac{(n+1)(x-4)}{n} \cdot \frac{n^3 + 1}{(n+1)^3 + 1} \right|.
   \]

4. **Take the Limit as \(n \to \infty\)**:
   Now, calculate the limit of this expression as \(n \to \infty\):
   \[
   L = \lim_{n \to \infty} \left| \frac{(n+1)(x-4)}{n} \cdot \frac{n^3 + 1}{(n+1)^3 + 1} \right|.
   \]
   
   As \(n \to \infty\), \(\frac{n+1}{n} \to 1\) and \(\frac{n^3 + 1}{(n+1)^3 + 1} \to 1\). So we get:
   \[
   L = |x - 4|.
   \]

5. **Determine Convergence Based on \(L\)**:
   - For absolute convergence, we need \(L < 1\), which means \(|x - 4| < 1\).
   - This simplifies to \(3 < x < 5\).

   Therefore, the series is absolutely convergent for \(3 < x < 5\).

### Conclusion
The series is absolutely convergent for \(3 < x < 5\) and divergent otherwise.

Would you like more details or have any questions?

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### Further Related Questions

1. What is the radius of convergence for this series?
2. How would the convergence test change if we used the root test instead of the ratio test?
3. For what values of \(x\) does the series converge conditionally?
4. How would we analyze the convergence if the series was centered around a different point?
5. What is the effect of the \(n\)-dependent numerator on the convergence?

### Tip
When applying the ratio test, simplify each term as much as possible before taking the limit to avoid unnecessary complexity.

Determine whether the series \( \sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1} \) is absolutely convergent or divergent using the ratio test.

Explore the convergence of the series \( \sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1} \) using the ratio test. This guide explains each step of the process, including setting up the ratio, finding the limit, and determining the range of values for absolute convergence.

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