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To determine the convergence or divergence of the series 

\[
\sum_{n=1}^{\infty} \frac{4n^2 + n - 1}{4n^4 + 8n^2 - 5}
\]

we can apply the Limit Comparison Test. For the Limit Comparison Test, we need to choose a comparison series \( b_n \) that resembles the original series at large values of \( n \).

### Step-by-Step Analysis

1. **Identify the dominant terms**: As \( n \) becomes very large, the dominant term in the numerator of \( \frac{4n^2 + n - 1}{4n^4 + 8n^2 - 5} \) is \( 4n^2 \), and the dominant term in the denominator is \( 4n^4 \). So, we can approximate the original series by the expression:

   \[
   \frac{4n^2}{4n^4} = \frac{1}{n^2}
   \]

   This suggests that we should try comparing with \( b_n = \frac{1}{n^2} \), as it matches the behavior of the original series at large values of \( n \).

2. **Set up the limit for comparison**: We use the Limit Comparison Test with \( b_n = \frac{1}{n^2} \). We calculate the limit:

   \[
   L = \lim_{n \to \infty} \frac{\frac{4n^2 + n - 1}{4n^4 + 8n^2 - 5}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(4n^2 + n - 1) \cdot n^2}{4n^4 + 8n^2 - 5}
   \]

3. **Simplify the expression**:

   Expanding and simplifying this limit gives:

   \[
   L = \lim_{n \to \infty} \frac{4n^4 + n^3 - n^2}{4n^4 + 8n^2 - 5}
   \]

   Dividing the numerator and the denominator by \( n^4 \), we get:

   \[
   L = \lim_{n \to \infty} \frac{4 + \frac{1}{n} - \frac{1}{n^2}}{4 + \frac{8}{n^2} - \frac{5}{n^4}}
   \]

   As \( n \to \infty \), the terms with \( \frac{1}{n}, \frac{1}{n^2}, \frac{8}{n^2} \), and \( \frac{5}{n^4} \) approach zero, leaving:

   \[
   L = \frac{4}{4} = 1
   \]

4. **Conclusion**: Since \( L = 1 \), which is a positive finite number, the Limit Comparison Test tells us that the convergence behavior of the original series is the same as that of \( \sum \frac{1}{n^2} \), which is a convergent \( p \)-series (since \( p = 2 > 1 \)).

Therefore, the original series **converges**.

### Answer to Multiple Choice
The correct choice for \( b_n \) is:
- \( b_n = \frac{1}{n^2} \), which gives \( L = 1 \).

Would you like additional details, or do you have any questions?

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### Related Questions

1. How does the Limit Comparison Test work when \( L = 0 \) or \( L = \infty \)?
2. What is a \( p \)-series, and when does it converge?
3. How does choosing different dominant terms affect the comparison series?
4. What other convergence tests could be used for this series?
5. How does the behavior of the terms in the numerator and denominator at large \( n \) affect the choice of \( b_n \)?

### Tip
Always identify the dominant terms in both the numerator and denominator when using the Limit Comparison Test to simplify the process.

Use the Limit Comparison Test to determine convergence or divergence for the series ∑ from n=1 to ∞ of (4n^2 + n - 1) / (4n^4 + 8n^2 - 5).

Determine the convergence of the series ∑ (4n^2 + n - 1) / (4n^4 + 8n^2 - 5) using the Limit Comparison Test. This problem includes identifying dominant terms and comparing to a p-Series for a detailed convergence analysis.

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