Chapter 8: Q 7. (page 704)

Find the interval of convergence of the power series
$\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$

Short Answer

Expert verified

The interval of convergence of the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$ is $(0, 4)$

Step by step solution

Given information

The power series is $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$

The ratio test for absolute convergence will be used to determine the convergence interval.

Let, the first assume $b_{k} = \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$

therefore $b_{k + 1} = \frac{(- 1)^{k + 1}}{(k + 1) 2^{k + 1}} (x - 2)^{k + 1}$

Implies that,

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{{(- 1)}^{k + 1}}{(k + 12^{k + 1}} {(x - 2)}^{k + 1}}{\frac{{(- 1)}^{k}}{k 2^{k}} {(x - 2)}^{k}}| = \lim_{k \to \infty} \frac{1}{2} \frac{k}{{(k + 1)}^{k}} |x - 2|$

The limit is $\frac{1}{2} | x - 2 |$

The ratio test for absolute convergence will be used to determine the convergence interval when $\frac{1}{2} | x - 2 | < 1$ that is $- 2 < x - 2 < 2$

As a result, we may write $- 2 < x - 2 a n d x - 2 < 2$

Implies that

$x > 0$ and $x < 4$

or $x \in (0, 4)$

Now, because the intervals are limited, we examine the series' behavior at the ends.

When $x = 0$

${\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}|}_{x = 0} = \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (0 - 2)^{k} = \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (- 1)^{k} 2^{k} = \sum_{k = 1}^{\infty} \frac{1}{k} (- 1)^{2 k} = \sum_{k = 1}^{\infty} \frac{1}{k}$

The series contains the diverging alternating multiple.

So, when $x = 4$

The alternating multiple is the outcome, and it converges.

The interval of convergence of the power series

The interval of convergence of the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$ is $(0, 4)$

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Find the interval of convergence of the power series
$\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$

Short Answer

Step by step solution

Given information

The ratio test for absolute convergence will be used to determine the convergence interval.

Now, because the intervals are limited, we examine the series' behavior at the ends.

The interval of convergence of the power series

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Statistics

Discrete Mathematics

Pure Maths

Geometry

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the interval of convergence of the power series ∑k=1∞(-1)kk2k(x-2)k

Short Answer

Step by step solution

Given information

The ratio test for absolute convergence will be used to determine the convergence interval.

Now, because the intervals are limited, we examine the series' behavior at the ends.

The interval of convergence of the power series

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Statistics

Discrete Mathematics

Pure Maths

Geometry

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the interval of convergence of the power series
$\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k 2^{k}} (x - 2)^{k}$