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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 10 - Sequences and Infinite Series문제 10.7.39
10장, 문제 10.7.39

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 k³ / √(k⁸ + 1)

검증된 단계별 안내
1
Identify the given series: \( \sum_{k=1}^{\infty} (-1)^k \frac{k^3}{\sqrt{k^8 + 1}} \). Notice it is an alternating series because of the factor \( (-1)^k \).
To determine absolute convergence, consider the absolute value of the terms: \( \left| \frac{k^3}{\sqrt{k^8 + 1}} \right| = \frac{k^3}{\sqrt{k^8 + 1}} \). Simplify this expression to understand its behavior as \( k \to \infty \).
Analyze the limit of the absolute value of the terms as \( k \to \infty \). Since \( \sqrt{k^8 + 1} = (k^8 + 1)^{1/2} \), rewrite it as \( k^4 \sqrt{1 + \frac{1}{k^8}} \). Then the term becomes \( \frac{k^3}{k^4 \sqrt{1 + \frac{1}{k^8}}} = \frac{1}{k \sqrt{1 + \frac{1}{k^8}}} \).
Evaluate the limit \( \lim_{k \to \infty} \frac{1}{k \sqrt{1 + \frac{1}{k^8}}} \). Since \( \sqrt{1 + \frac{1}{k^8}} \to 1 \), the limit behaves like \( \frac{1}{k} \), which goes to zero.
Since the terms behave like \( \frac{1}{k} \) for large \( k \), the absolute series resembles the harmonic series \( \sum \frac{1}{k} \), which diverges. Therefore, the series does not converge absolutely. Next, apply the Alternating Series Test by checking if the terms \( b_k = \frac{k^3}{\sqrt{k^8 + 1}} \) decrease monotonically to zero to determine conditional convergence.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Absolute and Conditional Convergence

Absolute convergence occurs when the series of absolute values converges, ensuring the original series converges regardless of term signs. Conditional convergence happens when the series converges but not absolutely, often due to alternating signs. Understanding these distinctions helps classify the behavior of infinite series.
추천 영상:
가이드 코스
07:51
Choosing a Convergence Test

Limit Comparison Test

The Limit Comparison Test compares a given series to a known benchmark series by examining the limit of their term ratios. If the limit is a positive finite number, both series share the same convergence behavior. This test is useful for series with complicated terms involving polynomials and roots.
추천 영상:
가이드 코스
07:45
Limit Comparison Test

Alternating Series Test

The Alternating Series Test determines convergence of series whose terms alternate in sign. It requires that the absolute value of terms decreases monotonically to zero. If these conditions hold, the series converges, which helps identify conditional convergence in alternating series.
추천 영상:
가이드 코스
10:54
Alternating Series Test
관련 실천
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Define the remainder of an infinite series.

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13–52. Limits of sequences

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


{tan⁻¹(10n⁄(10n + 4))}  

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21–42. Geometric series Evaluate each geometric series or state that it diverges.  


21.∑ (k = 0 to ∞) (1/4)ᵏ

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For what values of r does the sequence {rⁿ} converge? Diverge?

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Given the series ∑∞ₖ₌₁ k, evaluate the first four terms of its sequence of partial sums Sₙ = ∑ⁿₖ₌₁ k. 

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교과서 질문

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


∑ (k = 4 to ∞) (1 + cos²(k)) / (k − 3)

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