Q. 51 Use any convergence tests to det... [FREE SOLUTION]

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Chapter 7: Q. 51 (page 653)

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
${鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$

Short Answer

Expert verified

The series is absolutely convergent.

Step by step solution

Step 1. Given information.

Consider the given question,

${鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$

Step 2. Consider the general series.

The general term of the series ${鈭�}_{k = 1}^{鈭�} a_{k} = {鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$ is given below,

$a_{k} = \sin^{3} (\frac{1}{k})$

The limit comparison test states that for ${鈭�}_{k = 1}^{鈭�} a_{k}, {鈭�}_{k = 1}^{鈭�} b_{k}$ be two series with positive terms such that $0 鈮� a_{k} 鈮� b_{k}$ for every positive integer k. If the series ${鈭�}_{k = 1}^{鈭�} b_{k}$ converges, then the series ${鈭�}_{k = 1}^{鈭�} a_{k}$ .

The terms of the series ${鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$ are positive.

Step 3. Consider the series ∑k=1∞ sin31k.

The expression $\sin^{3} (\frac{1}{k})$ satisfies $|\sin^{3} (\frac{1}{k})| 鈮� \frac{1}{k^{3}}$ .

The series ${鈭�}_{k = 1}^{鈭�} b_{k}$ is given by ${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k^{3}}$ .

The series ${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k^{3}}$ is also convergent and converges absolutely.

Hence, the given series is absolutely convergent.

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91影视

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
${鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider the general series.

Step 3. Consider the series ∑k=1∞ sin31k.

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91影视

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.鈭�k=1鈭�sin31k

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider the general series.

Step 3. Consider the series ∑k=1∞ sin31k.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Decision Maths

Pure Maths

Logic and Functions

Calculus

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
${鈭�}_{k = 1}^{鈭�} \sin^{3} (\frac{1}{k})$