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

A p series other than ${鈭�}_{k = 1}^{鈭�} \frac{1}{k^{2}}$ you could use with comparison test to show that the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ converges.

Short Answer

Expert verified

It is convergent

Step by step solution

Step 1

Consider the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$

To determine p series that used to show that ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ is convergent

The terms of series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ are positive.

The series ${鈭�}_{k = 1}^{鈭�} b_{k}$ for the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ is

${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k^{3 / 2}}$

c) Step 2

The ratio $\lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}}$ is given

$\lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = \lim_{k 鈫� 鈭�} \frac{\frac{k - 1}{k^{3} + k + 1}}{\frac{1}{k^{3 / 2}}} = \lim_{k 鈫� 鈭�} \frac{k^{3 / 2} (k - 1)}{k^{3} + k + 1} = \lim_{k 鈫� 鈭�} \frac{k^{5 / 2} (1 + \frac{1}{k})}{k^{3} (1 + \frac{1}{k^{2}} + \frac{1}{k^{3}})} = \lim_{k 鈫� 鈭�} \frac{(1 + \frac{1}{k})}{k^{1 / 2} (1 + \frac{1}{k^{2}} + \frac{1}{k^{3}})} = 0$

c) Step 3

The value 0f $\lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = 0$

The series ${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k^{3 / 2}}$ is convergent by the p-series test

Then ${鈭�}_{k = 1}^{鈭�} a_{k}$ is also convergent

Then the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ is convergent and the p series is ${鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{k^{3 / 2}}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

A p series other than ${鈭�}_{k = 1}^{鈭�} \frac{1}{k^{2}}$ you could use with comparison test to show that the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ converges.

Short Answer

Step by step solution

Step 1

c) Step 2

c) Step 3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Statistics

Geometry

Applied Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

91影视

A p series other than 鈭�k=1鈭�1k2you could use with comparison test to show that the series 鈭�k=1鈭�k-1k3+k+1converges.

Short Answer

Step by step solution

Step 1

c) Step 2

c) Step 3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Statistics

Geometry

Applied Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

A p series other than ${鈭�}_{k = 1}^{鈭�} \frac{1}{k^{2}}$ you could use with comparison test to show that the series ${鈭�}_{k = 1}^{鈭�} \frac{k - 1}{k^{3} + k + 1}$ converges.