/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 In Exercises \(1-32,\) (a) find ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 6

In Exercises \(1-32,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=0}^{\infty}(2 x)^{n} $$

Short Answer

Expert verified
Radius: \( \frac{1}{2} \), converges for \(-\frac{1}{2} < x < \frac{1}{2}\). No absolute or conditional convergence at endpoints.

Step by step solution

01

Identify the Type of Series

The given series \( \sum_{n=0}^{\infty} (2x)^n \) is a geometric series. Geometric series have the form \( \sum_{n=0}^{\infty} ar^n \). Here, \( a = 1 \) and \( r = 2x \).
02

Determine Radius of Convergence

For a geometric series \( \sum_{n=0}^{\infty} ar^n \), it converges if the absolute value of the common ratio \( |r| < 1 \). Thus, \( |2x| < 1 \). Solving \( |2x| < 1 \), we have \( |x| < \frac{1}{2} \). Therefore, the radius of convergence \( R \) is \( \frac{1}{2} \).
03

Find Interval of Convergence

From \( |x| < \frac{1}{2} \), we derive the interval of convergence as \(-\frac{1}{2} < x < \frac{1}{2} \).
04

Test Endpoints for Absolute Convergence

Substitute \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \) into the series to see if it converges absolutely. For both endpoints, the geometric series becomes \( \sum (\pm 1)^n \), which diverges. Thus, there's no absolute convergence at the endpoints.
05

Evaluate Conditional Convergence

The series \( \sum_{n=0}^{\infty} (2x)^n \) is a geometric series, and geometric series can only converge absolutely. Therefore, there is no conditional convergence in this scenario.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a fascinating and straightforward type of series with a distinctive trait. It can be represented as \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the initial term and \( r \) is the common ratio that multiplies each term by the previous one. The given series \( \sum_{n=0}^{\infty} (2x)^n \) is a geometric series where \( a = 1 \) and \( r = 2x \). Geometric series are special because they have specific criteria for convergence: they will converge if the absolute value of the common ratio \( |r| < 1 \). This makes analyzing their convergence relatively simple.
Interval of Convergence
The interval of convergence refers to the set of \( x \) values for which a series converges. For geometric series, we rely on the condition \( |r| < 1 \) for convergence, where \( r \) is the common ratio. In the example series \( \sum_{n=0}^{\infty} (2x)^n \), we need \( |2x| < 1 \). Solving this inequality gives us \( |x| < \frac{1}{2} \). Thus, the interval of convergence is \(-\frac{1}{2} < x < \frac{1}{2} \). This is the range where the series smoothly converges without issues, excluding convergence at any endpoints for now.
Radius of Convergence
The radius of convergence \( R \) is a crucial measure in understanding the range over which a series can converge. It forms half of the interval width where the function of the series retains validity. For a geometric series, once we establish \( |r| < 1 \), we can directly determine \( R \). In the example \( (2x)^n \), solving \( |2x| < 1 \) leads to the conclusion that the radius of convergence is \( R = \frac{1}{2} \). This means that within a sphere of radius \( \frac{1}{2} \) centered at zero, the series will converge.
Absolute Convergence
Absolute convergence is a strong form of convergence for series and a reassuring property. A series converges absolutely if the series made by taking the absolute value of each of its terms also converges. For geometric series, such as \( \sum_{n=0}^{\infty} (2x)^n \), the concept simplifies. Since the series converges if \( |2x| < 1 \), testing for absolute convergence usually overlaps with testing for convergence. In this particular series when tested at the endpoints \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \), both cases result in divergence due to alternating incomplete sums like \( \sum (\pm 1)^n \). Thus, the series does not converge absolutely at the endpoints.
Conditional Convergence
Conditional convergence occurs when a series converges, but it does not converge absolutely. However, this notion doesn't hold for geometric series. These series are special in that their type of convergence is either a full absolute convergence or a divergence. The series \( \sum_{n=0}^{\infty} (2x)^n \) doesn't allow for conditional convergence at all. If a geometric series does not converge absolutely鈥攁nd for the case studied, since neither endpoint allows for absolute convergence, and the interior follows the absolute rule鈥攊t simply cannot show conditional convergence.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+x^{2} y=x, \quad y^{\prime}(0)=b \text { and } y(0)=a $$

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} $$

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ \left(1+x^{2}\right) y^{\prime}+2 x y=0, \quad y(0)=3 $$

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series鈥� convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{n}{n^{2}+1} $$

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x) .\) Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) , $$ \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.