To test this series for convergence

∞
∑
n
=
1
 
n
5
−
1
n
5
−
6
 

You could use the Limit Comparison Test, comparing it to the series   
∞
∑
n
=
1
 
1
n
p
  where  
p
 =
1
Incorrect

Completing the test, it shows the series:
Converges
Diverges
Correct
Question HelpQuestion 19:  Video1

To test the convergence of the series

\[
\sum_{n=1}^{\infty} \frac{n^5 - 1}{n^5 - 6},
\]

we can use the **Limit Comparison Test** with a known comparison series. The given series has terms that are close to \(1\) as \(n\) becomes large, specifically:

\[
\frac{n^5 - 1}{n^5 - 6} \approx 1 \text{ as } n \to \infty.
\]

For large \(n\), the dominant term in both the numerator and the denominator is \(n^5\). Therefore, for large \(n\), the terms of the series behave like:

\[
\frac{n^5 - 1}{n^5 - 6} \approx 1.
\]

### Step-by-step approach:

1. **Choose a comparison series:**
   Since the terms of the series approach \(1\), it is not immediately clear what comparison series to choose. However, if we look at the asymptotic behavior, the difference between the numerator and denominator suggests a comparison with the constant series \(1/n^p\).

2. **Limit Comparison Test:**
   Let’s compare the given series to the series:
   \[
   \sum_{n=1}^{\infty} 1.
   \]
   We use the limit comparison test with the series \(b_n = 1\), which diverges. The Limit Comparison Test requires calculating:

   \[
   \lim_{n \to \infty} \frac{\frac{n^5 - 1}{n^5 - 6}}{1} = \lim_{n \to \infty} \frac{n^5 - 1}{n^5 - 6} = 1.
   \]

3. **Conclusion from Limit Comparison Test:**
   Since the limit is a positive finite number (1), and the comparison series \(\sum_{n=1}^{\infty} 1\) diverges, by the **Limit Comparison Test**, the original series also **diverges**.

### Correct conclusion:
The series **diverges**.

---

Would you like further details on any part of this process or to explore related topics? 

Here are 5 related questions to expand your understanding:

1. How does the Limit Comparison Test differ from the Direct Comparison Test?
2. Why does the series \(\sum_{n=1}^{\infty} 1/n^p\) converge when \(p > 1\)?
3. What other comparison tests can be used to determine convergence of series?
4. How can you recognize which series to compare using the Limit Comparison Test?
5. What are the conditions for using the Ratio Test instead of the Limit Comparison Test?

**Tip:** Always check the behavior of the series for large \(n\), as the asymptotic behavior often gives insight into convergence or divergence.

To test this series for convergence ∞ ∑ n=1 (n^5 - 1) / (n^5 - 6). You could use the Limit Comparison Test, comparing it to the series ∞ ∑ n=1 (1 / n^p) where p = 1. Completing the test, it shows the series: Converges or Diverges?

Understand how to test for the convergence of the series ∑ (n^5 - 1) / (n^5 - 6) using the Limit Comparison Test. Learn step-by-step how to apply the Limit Comparison Test and why the series diverges. This problem covers advanced calculus concepts, suitable for college-level or advanced high school math students.

Math Problem Statement

∞ ∑ n

You could use the Limit Comparison Test, comparing it to the series
∞ ∑ n

1 n p where
p

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