Math Problem Statement

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).

Solution

To determine the convergence or divergence of the series

n=14n2+n14n4+8n25\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 bnb_n that resembles the original series at large values of nn.

Step-by-Step Analysis

  1. Identify the dominant terms: As nn becomes very large, the dominant term in the numerator of 4n2+n14n4+8n25\frac{4n^2 + n - 1}{4n^4 + 8n^2 - 5} is 4n24n^2, and the dominant term in the denominator is 4n44n^4. So, we can approximate the original series by the expression:

    4n24n4=1n2\frac{4n^2}{4n^4} = \frac{1}{n^2}

    This suggests that we should try comparing with bn=1n2b_n = \frac{1}{n^2}, as it matches the behavior of the original series at large values of nn.

  2. Set up the limit for comparison: We use the Limit Comparison Test with bn=1n2b_n = \frac{1}{n^2}. We calculate the limit:

    L=limn4n2+n14n4+8n251n2=limn(4n2+n1)n24n4+8n25L = \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=limn4n4+n3n24n4+8n25L = \lim_{n \to \infty} \frac{4n^4 + n^3 - n^2}{4n^4 + 8n^2 - 5}

    Dividing the numerator and the denominator by n4n^4, we get:

    L=limn4+1n1n24+8n25n4L = \lim_{n \to \infty} \frac{4 + \frac{1}{n} - \frac{1}{n^2}}{4 + \frac{8}{n^2} - \frac{5}{n^4}}

    As nn \to \infty, the terms with 1n,1n2,8n2\frac{1}{n}, \frac{1}{n^2}, \frac{8}{n^2}, and 5n4\frac{5}{n^4} approach zero, leaving:

    L=44=1L = \frac{4}{4} = 1

  4. Conclusion: Since L=1L = 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 1n2\sum \frac{1}{n^2}, which is a convergent pp-series (since p=2>1p = 2 > 1).

Therefore, the original series converges.

Answer to Multiple Choice

The correct choice for bnb_n is:

  • bn=1n2b_n = \frac{1}{n^2}, which gives L=1L = 1.

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


Related Questions

  1. How does the Limit Comparison Test work when L=0L = 0 or L=L = \infty?
  2. What is a pp-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 nn affect the choice of bnb_n?

Tip

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

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Limit Comparison Test
Asymptotic Comparison
p-Series

Formulas

L = lim (n → ∞) (a_n / b_n)
p-Series: ∑ (1/n^p)

Theorems

Limit Comparison Test
p-Series Test

Suitable Grade Level

College Calculus