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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.R.47
Chapter 10, Problem 10.R.47

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)(7 + sin k) / k²

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Identify the given series: \( \sum_{k=1}^{\infty} \frac{7 + \sin k}{k^2} \). We want to determine if this series converges or diverges.
Note that \( \sin k \) is bounded between -1 and 1, so the numerator \( 7 + \sin k \) is bounded between 6 and 8. This means the terms behave roughly like \( \frac{\text{constant}}{k^2} \) for large \( k \).
Recall the p-series test: \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) converges if \( p > 1 \). Here, since the denominator is \( k^2 \), which corresponds to \( p = 2 > 1 \), the series \( \sum \frac{1}{k^2} \) converges.
Use the Comparison Test by comparing \( \frac{7 + \sin k}{k^2} \) with \( \frac{8}{k^2} \) (since 8 is an upper bound for the numerator). Since \( \sum \frac{8}{k^2} \) converges, and \( 0 \leq \frac{7 + \sin k}{k^2} \leq \frac{8}{k^2} \), the original series converges by the Comparison Test.
Conclude that the series \( \sum_{k=1}^{\infty} \frac{7 + \sin k}{k^2} \) converges absolutely because it is bounded by a convergent p-series with \( p = 2 \).

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Key Concepts

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

Convergence and Divergence of Infinite Series

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely. If the sum does not approach a finite value, the series diverges. Understanding this distinction is fundamental to analyzing series behavior.
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Convergence of an Infinite Series

Comparison Test for Series Convergence

The Comparison Test involves comparing a given series to a second series whose convergence behavior is known. If the terms of the given series are smaller than those of a convergent series, it also converges. This test is useful when terms are positive and can be bounded.
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Direct Comparison Test

Behavior of the p-Series

A p-series has the form ∑ 1/k^p and converges if and only if p > 1. Since the given series has terms involving 1/k², recognizing this helps determine convergence by comparing to a known convergent p-series with p=2.
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P-Series and Harmonic Series
Related Practice
Textbook Question

Finding steady states using infinite series Solve Exercise 40 by expressing the amount of aspirin in your blood as a geometric series and evaluating the series.

Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)5ᵏ / 2²ᵏ⁺¹

Textbook Question

12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.

aₙ = (–1)ⁿ (3n³ + 4n) / (6n³ + 5)

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Textbook Question

{Use of Tech} Error in a finite alternating sum

How many terms of the series ∑ (from k = 1 to ∞)(−1)ᵏ⁺¹ / k⁴ must be summed to ensure that the approximation error is less than 10⁻⁸?

Textbook Question

Explain why or why not

Determine whether the following statements are true and give an explanation or counterexample.

a.The terms of the sequence {aₙ} increase in magnitude, so the limit of the sequence does not exist.

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Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)2ᵏ / eᵏ