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Find the interval, *I*, of convergence of the series. (Enter your answer using interval notation.)

The given series is:

\[
\sum_{n=2}^{\infty} \frac{(x+5)^n}{5^n \ln(n)}
\]

To find the interval of convergence, we can use the **Ratio Test**. The Ratio Test states that for a series \(\sum a_n\), the series converges if the limit

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

satisfies \(L < 1\).

### Step 1: Define the general term \(a_n\)

The general term of the series is:

\[
a_n = \frac{(x+5)^n}{5^n \ln(n)}
\]

### Step 2: Compute the ratio \(\frac{a_{n+1}}{a_n}\)

\[
\frac{a_{n+1}}{a_n} = \frac{\frac{(x+5)^{n+1}}{5^{n+1} \ln(n+1)}}{\frac{(x+5)^n}{5^n \ln(n)}} = \frac{(x+5)^{n+1} \cdot 5^n \cdot \ln(n)}{(x+5)^n \cdot 5^{n+1} \cdot \ln(n+1)}
\]

Simplifying:

\[
\frac{a_{n+1}}{a_n} = \frac{(x+5) \cdot \ln(n)}{5 \cdot \ln(n+1)}
\]

### Step 3: Compute the limit \(L\)

\[
L = \lim_{n \to \infty} \left| \frac{(x+5) \ln(n)}{5 \ln(n+1)} \right|
\]

For large \(n\), \(\ln(n+1)\) behaves like \(\ln(n)\). Therefore:

\[
L = \lim_{n \to \infty} \left| \frac{(x+5)}{5} \cdot \frac{\ln(n)}{\ln(n+1)} \right| = \left| \frac{x+5}{5} \right| \cdot \lim_{n \to \infty} \frac{\ln(n)}{\ln(n+1)}
\]

Since \(\lim_{n \to \infty} \frac{\ln(n)}{\ln(n+1)} = 1\):

\[
L = \left| \frac{x+5}{5} \right|
\]

### Step 4: Apply the Ratio Test

For the series to converge, we need:

\[
\left| \frac{x+5}{5} \right| < 1
\]

This implies:

\[
-1 < \frac{x+5}{5} < 1
\]

Multiplying through by 5:

\[
-5 < x+5 < 5
\]

Subtracting 5 from all parts:

\[
-10 < x < 0
\]

### Step 5: Test the endpoints

1. **At \(x = -10\)**:
   \[
   \frac{(x+5)^n}{5^n \ln(n)} = \frac{(-10+5)^n}{5^n \ln(n)} = \frac{(-5)^n}{5^n \ln(n)} = \frac{(-1)^n}{\ln(n)}
   \]
   The series \(\sum \frac{(-1)^n}{\ln(n)}\) is an alternating series where the terms approach 0 as \(n\) increases, but \(\frac{1}{\ln(n)}\) does not decrease fast enough to ensure convergence by the Alternating Series Test or the Integral Test. Therefore, it diverges.

2. **At \(x = 0\)**:
   \[
   \frac{(x+5)^n}{5^n \ln(n)} = \frac{5^n}{5^n \ln(n)} = \frac{1}{\ln(n)}
   \]
   The series \(\sum \frac{1}{\ln(n)}\) also diverges because it resembles a harmonic series, which diverges.

### Conclusion

The series converges for \(x \in (-10, 0)\).

Thus, the interval of convergence is:

\[
I = (-10, 0)
\]

Would you like further details or have any questions?

---

Here are some related questions you might consider:

1. What is the Ratio Test and how does it determine convergence?
2. How can the Root Test be applied to determine convergence?
3. What are the endpoints of convergence and how are they tested?
4. How does the Alternating Series Test work for testing convergence?
5. What is the behavior of logarithmic functions at large values of \(n\)?
6. What happens if the series has absolute convergence within the interval?
7. How does the Integral Test help in determining the convergence of a series?
8. What are some practical examples of series that involve logarithmic terms?

**Tip:** When checking the endpoints of convergence, always consider applying tests like the Alternating Series Test or Integral Test to determine if they are included in the interval.

Learn how to determine the interval of convergence of a series using the Ratio Test with a detailed step-by-step solution. This problem covers series convergence, the Ratio Test, and interval notation. Suitable for advanced undergraduate students and beyond.

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