Q.2.b) As a p- series you could use a c... [FREE SOLUTION]

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Chapter 7: Q.2.b) (page 631)

As a p- series you could use a comparison to show that the series ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ diverges.

Short Answer

Expert verified

The ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ is divergent

Step by step solution

The objective is to find the p- series that is used to show that the series ∑k=1∞sin1k is convergent

The comparison test is used to determine the convergence or divergence of the series

It states that ${鈭�}_{k = 1}^{鈭�} a_{k}$ and ${鈭�}_{k = 1}^{鈭�} b_{k}$ 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 ${鈭�}_{k = 1}^{鈭�} a_{k}$ also convergences

According to the given data

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

The expression of $\sin (\frac{1}{k})$ follows inequality

$\sin (\frac{1}{k}) 鈮� \frac{1}{k}$

The series ${鈭�}_{k = 1}^{鈭�} b_{k}$ for the series ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ is given by

${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k}$

By concluding

The series ${鈭�}_{k = 1}^{鈭�} b_{k}$ = ${鈭�}_{k = 1}^{鈭�} \frac{1}{k}$ is divergent by the p- series

Therefore the ${鈭�}_{k = 1}^{鈭�} a_{k}$ is also divergent

Hence fore the ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ is divergent and p-series is ${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k}$

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91影视

As a p- series you could use a comparison to show that the series ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ diverges.

Short Answer

Step by step solution

The objective is to find the p- series that is used to show that the series ∑k=1∞sin1k is convergent

According to the given data

By concluding

One App. One Place for Learning.

Most popular questions from this chapter

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91影视

As a p- series you could use a comparison to show that the series鈭�k=1鈭�sin1kdiverges.

Short Answer

Step by step solution

The objective is to find the p- series that is used to show that the series ∑k=1∞sin1k is convergent

According to the given data

By concluding

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Applied Mathematics

Decision Maths

Pure Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

As a p- series you could use a comparison to show that the series ${鈭�}_{k = 1}^{鈭�} \sin (\frac{1}{k})$ diverges.