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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 10.5.23
Chapter 10, Problem 10.5.23

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.


∑ (k = 1 to ∞) sin(1 / k) / k²

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First, identify the general term of the series: \(a_k = \frac{\sin(1/k)}{k^2}\).
Recall that for large \(k\), \(\sin(1/k)\) behaves approximately like \(1/k\) because \(\sin x \approx x\) when \(x\) is close to 0.
Use this approximation to compare \(a_k\) with a simpler series term: \(b_k = \frac{1/k}{k^2} = \frac{1}{k^3}\).
Since \(\sum_{k=1}^\infty \frac{1}{k^3}\) is a p-series with \(p=3 > 1\), it converges.
Apply the Limit Comparison Test by computing \(\lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\sin(1/k)/k^2}{1/k^3} = \lim_{k \to \infty} \frac{\sin(1/k)}{1/k}\). Since this limit is finite and nonzero, both series share the same convergence behavior, so the original series converges.

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Key Concepts

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

Comparison Test

The Comparison Test determines the convergence of a series by comparing it to a second series with known behavior. If the terms of the given series are smaller than those of a convergent series with positive terms, then the given series also converges. Conversely, if the terms are larger than those of a divergent series, the given series diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares two series by taking the limit of the ratio of their terms. If this limit is a positive finite number, both series either converge or diverge together. This test is useful when direct comparison is difficult but the terms behave similarly for large indices.
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Limit Comparison Test

Behavior of sin(1/k) for large k

For large values of k, sin(1/k) can be approximated by its argument 1/k because sin(x) ≈ x when x is close to zero. This approximation helps simplify the given series terms to roughly 1/(k³), aiding in the application of comparison tests with p-series.
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Newton's Law of Cooling Example 5
Related Practice
Textbook Question

Explain why the magnitude of the remainder in an alternating series (with terms that are nonincreasing in magnitude) is less than or equal to the magnitude of the first neglected term.

Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.


∑ (from k = 0 to ∞)3k / ∜(k⁴ + 3)

Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{tan⁻¹(n)}

Textbook Question

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) (−1)ᵏ⁺¹ · (k!) / (kᵏ)(Hint: Show that k! / kᵏ ≤ 2 / k², for k ≥ 3.)

Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{ⁿ√(e³ⁿ⁺⁴)} 

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Textbook Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) (−1)ᵏ / k⁰.⁹⁹