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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.7.87a
Chapter 10, Problem 10.7.87a

Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
a. If the Limit Comparison Test can be applied successfully to a given series with a certain comparison series, the Comparison Test also works with the same comparison series.

Verified step by step guidance
1
Recall the definitions of the Comparison Test and the Limit Comparison Test for series convergence: The Comparison Test requires that the terms of the given series and the comparison series satisfy an inequality for all sufficiently large indices, while the Limit Comparison Test uses the limit of the ratio of their terms as the index approaches infinity.
For the Comparison Test, you need to find a comparison series \( \sum b_n \) such that either \( 0 \leq a_n \leq b_n \) or \( 0 \leq b_n \leq a_n \) for all \( n \) beyond some index, and then use the known convergence or divergence of \( \sum b_n \) to conclude about \( \sum a_n \).
For the Limit Comparison Test, you compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \), where \( c \) is a finite positive number. If this limit exists and is positive and finite, then both series either converge or diverge together.
Consider that the Limit Comparison Test can succeed even if the inequality required for the Comparison Test does not hold for all large \( n \). This means the Comparison Test might fail to apply with the same comparison series even though the Limit Comparison Test works.
Therefore, the statement is false: the successful application of the Limit Comparison Test with a certain comparison series does not guarantee that the Comparison Test will also work with the same comparison series. A counterexample can be constructed where the limit of the ratio exists and is positive, but the inequality needed for the Comparison Test fails.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limit Comparison Test

The Limit Comparison Test determines the convergence or divergence of a series by comparing it to a second series with known behavior. It involves taking the limit of the ratio of their terms; if the limit is a positive finite number, both series either converge or diverge together.
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07:45
Limit Comparison Test

Comparison Test

The Comparison Test assesses a series' convergence by directly comparing its terms to those of a known convergent or divergent series. If the terms of the original series are smaller than a convergent series, it converges; if larger than a divergent series, it diverges.
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Direct Comparison Test

Differences Between Limit Comparison and Comparison Tests

While both tests compare series, the Comparison Test requires strict inequalities between terms, which can be restrictive. The Limit Comparison Test is more flexible, relying on the limit of term ratios, so it can succeed even when direct inequalities needed for the Comparison Test do not hold.
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Limit Comparison Test
Related Practice
Textbook Question

18–20. Evaluating geometric series two ways Evaluate each geometric series two ways.


a. Find the nth partial sum Sₙ of the series and evaluate lim (as n → ∞) Sₙ.


∑ (k = 0 to ∞) (–2/7)ᵏ

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Textbook Question

57–60. Heights of bouncing balls A ball is thrown upward to a height of hₒ meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let hₙ be the height after the nth bounce. Consider the following values of hₒ and r.


a. Find the first four terms of the sequence of heights {hₙ}.


h₀ = 30,r = 0.25

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Textbook Question

39–40. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, use Theorem 10.13 to complete the following.


a. Use Sₙ to estimate the sum of the series.


39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Textbook Question

Find the first term a and the ratio r of each geometric series.


a. ∑ k = 0 to ∞(2/3) × (1/5)ᵏ

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. Suppose 0 < aₖ < bₖ. If ∑ (k = 1 to ∞) aₖ converges, then ∑ (k = 1 to ∞) bₖ converges.

Textbook Question

57–60. Heights of bouncing balls A ball is thrown upward to a height of hₒ meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let hₙ be the height after the nth bounce. Consider the following values of hₒ and r.


a. Find the first four terms of the sequence of heights {hₙ}.


h₀ = 20,r = 0.5