Math Problem Statement

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

Solution

To test the convergence of the series

n=1n51n56,\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 11 as nn becomes large, specifically:

n51n561 as n.\frac{n^5 - 1}{n^5 - 6} \approx 1 \text{ as } n \to \infty.

For large nn, the dominant term in both the numerator and the denominator is n5n^5. Therefore, for large nn, the terms of the series behave like:

n51n561.\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 11, 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/np1/n^p.

  2. Limit Comparison Test: Let’s compare the given series to the series: n=11.\sum_{n=1}^{\infty} 1. We use the limit comparison test with the series bn=1b_n = 1, which diverges. The Limit Comparison Test requires calculating:

    limnn51n561=limnn51n56=1.\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 n=11\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 n=11/np\sum_{n=1}^{\infty} 1/n^p converge when p>1p > 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 nn, as the asymptotic behavior often gives insight into convergence or divergence.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Limit Comparison Test
Asymptotic Behavior

Formulas

Limit Comparison Test: lim(n→∞) (a_n / b_n)
Asymptotic approximation of series terms

Theorems

Limit Comparison Test
Divergence of Harmonic Series

Suitable Grade Level

Undergraduate Calculus or Advanced High School Math