/*! 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 22 (a) find the series' radius and ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$\sum_{n=1}^{\infty} \frac{(-1)^{n} 3^{2 n}(x-2)^{n}}{3 n}$$

Short Answer

Expert verified
Radius: \(\frac{1}{9}\), Interval: \(\left(\frac{17}{9}, \frac{19}{9}\right)\). Absolute convergence: \(x \in \left(\frac{17}{9}, \frac{19}{9}\right)\). Conditional: endpoints.

Step by step solution

01

Identify the Series

The given series is \(\sum_{n=1}^{\infty} \frac{(-1)^{n} 3^{2 n}(x-2)^{n}}{3 n}\). This is a series with alternating terms due to \((-1)^{n}\), and it involves powers of \((x-2)\). We suspect that this resembles a power series.
02

Determine Power Series Form

Rewrite the given series in a form more recognizable as a power series: \(\sum_{n=1}^{\infty} (-1)^{n} \cdot \frac{3^{2n}}{3n} \cdot (x-2)^{n}\). The series is in the form of \(\sum_{n=1}^{\infty} a_n (x-2)^n\), where \(a_n = \frac{(-1)^n 3^{2n}}{3n}\).
03

Find the Radius of Convergence Using Ratio Test

For the series \(\sum a_n (x-2)^n\), we use the Ratio Test to find the radius of convergence. Compute:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \cdot \frac{3^{2(n+1)}}{3(n+1)}}{(-1)^{n} \cdot \frac{3^{2n}}{3n}} \right| = \lim_{n \to \infty} \frac{3^2 \cdot n}{3(n+1)} = 9 \cdot \lim_{n \to \infty} \frac{n}{n+1} = 9.\]Since the limit of \(\frac{n}{n+1}\) equals 1 as \(n\) approaches infinity, the reciprocal of this result gives the radius: \(R = \frac{1}{9}\).
04

Find the Interval of Convergence

With \(R = \frac{1}{9}\), the interval of convergence is centered around \(x = 2\). Therefore, the interval is \((2 - \frac{1}{9}, 2 + \frac{1}{9})\), or equivalently, \(\left(\frac{17}{9}, \frac{19}{9}\right)\).Examine the endpoints separately for convergence of the series.
05

Determine Absolute Convergence

A series converges absolutely if the series formed by taking the absolute values of its terms converges. Consider:\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^{n} 3^{2 n} (x-2)^n}{3n} \right| = \sum_{n=1}^{\infty} \frac{3^{2n} |x-2|^n}{3n}.\]This is a form suitable for another ratio test, confirming convergence when \(|x-2| < R\), which is already established to be \(x \in \left(\frac{17}{9}, \frac{19}{9}\right)\).
06

Determine Conditional Convergence

For conditional convergence, the series must converge but not absolutely at the endpoints. Use tests like the Alternating Series Test to evaluate convergence at \(x = \frac{17}{9}\) and \(x = \frac{19}{9}\), checking if series terms decrease in magnitude to zero.
07

Endpoint Testing

Evaluating separately at the endpoints:- For \(x = \frac{17}{9}\), the series becomes \(\sum (-1)^n \cdot \frac{1}{3n}\), which converges by the Alternating Series Test (terms decrease to zero).- Similarly, for \(x = \frac{19}{9}\), the series \(\sum (-1)^n \cdot \frac{1}{3n}\) also converges for the same reason.Since the series converges conditionally at both endpoints, those points are included in the convergence interval for conditional convergence.

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 91Ó°ÊÓ!

Key Concepts

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

Alternating Series
In mathematical series, an alternating series is one where the sign of each term alternates between positive and negative. This feature is characterized by the presence of \((-1)^n\) in the terms of the series. Such sequences are important because they often converge even when similar non-alternating series do not.

For the given series, \((-1)^n\) indicates that the series is indeed alternating. The Alternating Series Test is a common method to determine convergence of alternating series. This test states that if the absolute value of the terms is decreasing and approaching zero, then the series converges. When considering this series, it is crucial to apply this test to check convergence, especially at specific points such as the endpoints of the interval of convergence.
Radius of Convergence
The radius of convergence of a power series is a measure of the interval over which the series converges. It is found using the Ratio Test or the Root Test, both of which analyze the behavior of the terms as the series progresses.

For our original problem, we identified the radius of convergence by using the Ratio Test. By computing \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 9, \) where \(a_n\) represents the coefficient series, the radius of convergence \(R\) was determined to be \(\frac{1}{9}\). This radius \(R\) essentially dictates that the series will converge for values of \(x\) such that \(|x-2| < \frac{1}{9}\). This insight proves critical when exploring the boundaries of convergence.
Endpoint Convergence
Endpoint convergence refers to evaluating whether a series converges at the endpoints of its interval of convergence. Once the interval has been established using the radius of convergence, it's necessary to test these endpoints separately.

In our series, the interval is \(\left(\frac{17}{9}, \frac{19}{9}\right)\). At these endpoints, the nature of convergence is determined by plugging in the endpoint values into the series and employing convergence tests like the Alternating Series Test. In this specific problem, both endpoints showed conditional convergence when tested, meaning the series converges, but not absolutely, at these points. Such detailed checking is crucial for a complete understanding of the series behavior across its interval.
Power Series
A power series is a series of the form \(\sum_{n=0}^{\infty} a_n (x-c)^n\), where \(a_n\) are the coefficients and \(c\) is the center. Such series can approximate functions over specific intervals, known as intervals of convergence, dictated by the radius of convergence.

In our exercise, the series was a power series centered at \(x=2\), with terms involving powers of \(x-2\). This framework fits the model of a power series, allowing us to use various tests to assess convergence. Understanding this concept is key because it facilitates the use of powerful mathematical tools like the Ratio Test to examine convergence within the radius and scrutinize endpoints separately.
Ratio Test
The Ratio Test is a valuable method for determining convergence of a series. It can particularly help with power series to identify the radius of convergence. The test examines the absolute value of the ratio of consecutive terms \(\left| \frac{a_{n+1}}{a_n} \right|\) and its limit as \(n\) approaches infinity.

Applying the Ratio Test in our given series, the calculated limit was 9. This indicated that for the series to converge, \(|x-2| < \frac{1}{9}\). The Ratio Test can both verify convergence of the series over the interval and guide us in establishing the convergence radius, which is crucial for predicting where the entire series will reliably reflect function behavior. The understanding and usage of the Ratio Test is pivotal in working out the nature of power series.

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

a. Find the interval of convergence of the power series $$\sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n}$$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}.$$to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

a. Series for \(\sinh ^{-1} x \quad\) Find the first four nonzero terms of the Taylor series for $$ \sinh ^{-1} x=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ b. Use the first three terms of the series in part (a) to estimate \(\sinh ^{-1} 0.25 .\) Give an upper bound for the magnitude of the estimation error.

The English mathematician Wallis discovered the formula $$\frac{\pi}{4}=\frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot \cdots}$$ Find \(\pi\) to two decimal places with this formula.

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

See all solutions

Recommended explanations on Math Textbooks

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.