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

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

we'll apply the Ratio Test.

### Step 1: Set Up 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:
- If \(L < 1\), the series converges absolutely.
- If \(L > 1\), the series diverges.
- If \(L = 1\), the test is inconclusive.

Here, our \(a_n\) is given by:

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

### Step 2: Find \(a_{n+1}\) and Set Up the Limit

The term \(a_{n+1}\) will be:

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

Now, consider the ratio \(\left| \frac{a_{n+1}}{a_n} \right|\):

\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)(x - 4)^{n+1}}{(n+1)^3 + 1} \cdot \frac{n^3 + 1}{n (x - 4)^n} \right|.
\]

Simplifying this expression:

\[
= \left| \frac{(n+1)}{n} \cdot \frac{x - 4}{1 + \frac{1}{(n+1)^3}} \right|.
\]

As \(n \to \infty\), \(\frac{n+1}{n} \to 1\) and \(\frac{1}{(n+1)^3} \to 0\), so this expression approaches \(|x - 4|\).

Thus,

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |x - 4|.
\]

### Step 3: Apply the Ratio Test for Convergence

For convergence of the series, we need

\[
|x - 4| < 1.
\]

This gives us the radius of convergence \(R = 1\).

### Step 4: Determine the Interval of Convergence

The interval of convergence will be centered at \(x = 4\), with radius \(1\). So we consider the interval:

\[
3 < x < 5.
\]

To fully determine the interval of convergence, we need to check the endpoints \(x = 3\) and \(x = 5\):

1. **At \(x = 3\):** The series becomes
   \[
   \sum_{n=1}^{\infty} \frac{n (-1)^n}{n^3 + 1}.
   \]
   This series does not converge because the terms \(\frac{n}{n^3 + 1} \to 0\) very slowly and do not meet the criteria for absolute or conditional convergence.

2. **At \(x = 5\):** The series becomes
   \[
   \sum_{n=1}^{\infty} \frac{n (1)^n}{n^3 + 1}.
   \]
   Similarly, this does not converge for the same reason as at \(x = 3\).

Thus, the series converges only within the open interval \( (3, 5) \).

### Final Answer

- **Radius of Convergence, \(R\):** \(1\).
- **Interval of Convergence:** \( (3, 5) \).

Would you like further details on any part of this solution?

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**Follow-up Questions:**

1. How does the Ratio Test determine convergence for series with variable terms?
2. What other convergence tests could be used for similar series?
3. Why is it important to check endpoint convergence separately?
4. How does absolute convergence differ from conditional convergence?
5. Can the Ratio Test be applied to all types of series?

**Tip:** Always check endpoint values separately when finding the interval of convergence with the Ratio Test, as the test itself is inconclusive at the boundaries.

Find the radius of convergence and the interval of convergence of the given series using ratio test: \( \sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3+1} \).

Determine the radius and interval of convergence for the series \( \sum \frac{n(x-4)^n}{n^3+1} \) by applying the Ratio Test. This solution involves analyzing series convergence criteria and calculating both radius and interval of convergence, useful for students at the undergraduate level.

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