Chapter 9: Problem 7
Find the convergence set for the given power series. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n}\)
Short Answer
Expert verified
The convergence set for the series is \( 1 < x \leq 3 \).
Step by step solution
01
Identify the Power Series and Use Ratio Test
The given power series is \( \sum_{n=1}^{\infty} (-1)^{n} \frac{(x-2)^{n}}{n} \). We will use the Ratio Test, which is used to find the radius of convergence by calculating \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = (-1)^n \frac{(x-2)^n}{n} \).
02
Compute the Ratio of Successive Terms
Compute \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} \frac{(x-2)^{n+1}}{n+1}}{(-1)^n \frac{(x-2)^n}{n}} \right| = \left| \frac{(x-2)^{n+1}}{(n+1)} \cdot \frac{n}{(x-2)^n} \right| = \left| (x-2) \cdot \frac{n}{n+1} \right|.\) As \( n \to \infty \), \( \frac{n}{n+1} \to 1 \).
03
Apply the Ratio Test
The limit from the ratio test becomes \( \left|(x-2)\right| \cdot 1 = \left|(x-2)\right| \). For the series to converge, \( \left|(x-2)\right| < 1 \), which implies \( -1 < x-2 < 1 \).
04
Simplify the Inequality for x
From \( -1 < x-2 < 1 \), add 2 to each part of the inequality: \( 1 < x < 3 \). Thus, the radius of convergence is 1, centered at \( x = 2 \).
05
Test the Endpoints for Convergence
Substitute the endpoints \( x = 1 \) and \( x = 3 \) into the original series:- For \( x = 1 \), the series becomes \( \sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n} = -\sum_{n=1}^{\infty} \frac{1}{n} \), which diverges by the harmonic series.- For \( x = 3 \), the series becomes \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{n} \), which converges by the alternating series test.
06
Conclusion on Convergence Set
The convergence set is \( 1 < x \leq 3 \), because the series converges at \( x = 3 \) and diverges at \( x = 1 \).
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.
Ratio Test
The Ratio Test is a useful tool in analyzing the convergence of a power series. In this context, it helps us find the radius of convergence. For a series \( \sum_{n=1}^{\infty} a_n \), the Ratio Test involves the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges; if it's greater than 1, the series diverges. However, if the limit equals 1, the test is inconclusive. In the original exercise, the Ratio Test was applied to the series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n} \). By computing \( \left| \frac{a_{n+1}}{a_n} \right| \), we found that the limit equals \( \left|(x-2)\right| \). For convergence, \( \left|(x-2)\right| < 1 \) must hold, giving us the condition \( 1 < x < 3 \). With this, we identified the basic interval of convergence.
Alternating Series
An alternating series is one where the terms alternate between positive and negative. Such a series often has the form \( \sum_{n=1}^{\infty} (-1)^{n} b_n \), where \( b_n > 0 \). Alternating series converge under specific conditions, detailed in the Alternating Series Test. This test requires two things:
- The absolute values of the terms must be decreasing: \( b_{n+1} \leq b_n \)
- The limit of the terms must approach zero: \( \lim_{n \to \infty} b_n = 0 \)
Radius of Convergence
The radius of convergence is a crucial concept that informs us over what range of values the power series converges. It is often found using the Ratio Test. If the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| x - c \right| \), we solve \( \left| x - c \right| < R \) to find the radius \( R \).In the given series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n} \), the center \( c \) is 2. From the Ratio Test, we derived \( \left| x - 2 \right| < 1 \), thus confirming the radius of convergence as 1. The series converges for \( x \) values within a distance of 1 from the center, accurately between \( 1 < x < 3 \). Remember: while this tells us where convergence is possible, we must check the endpoints separately to know exactly at which points, if any, the series converges.