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Chapter 11: Problem 4

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? $$ \sum_{n=1}^{\infty} \frac{(3 x-2)^{n}}{n} $$

Short Answer

Expert verified
Radius of convergence: \( \frac{1}{3} \), converges absolutely for \( \frac{1}{3} < x < 1 \), conditionally at \( x = \frac{1}{3} \).

Step by step solution

01

Identifying the Series Structure

The given series is \( \sum_{n=1}^{\infty} \frac{(3x-2)^n}{n} \), which is similar to a geometric series but does not have the same form. Instead, it resembles a power series centered at \(x = \frac{2}{3}\).
02

Applying the Ratio Test for Convergence

To find the radius of convergence, apply the Ratio Test: Calculate \( \lim_{n \to \infty} \left| \frac{(3x-2)^{n+1}}{n+1} \cdot \frac{n}{(3x-2)^n} \right| \). Simplifying gives \( |3x-2| \cdot \lim_{n \to \infty} \frac{n}{n+1} = |3x-2| \). The series converges if \( |3x-2| < 1 \).
03

Finding the Radius and Interval of Convergence

The inequality \( |3x-2| < 1 \) simplifies to \(-1 < 3x - 2 < 1\). Solving this gives the interval of convergence \(\frac{1}{3} < x < 1\). The radius of convergence is therefore \( \frac{1}{3} \).
04

Checking Endpoint Convergence

Evaluate the series at the endpoints \( x = \frac{1}{3} \) and \( x = 1 \). At \( x = \frac{1}{3} \), the series becomes \( \sum_{n=1}^{\infty} 0^n / n = 0 \), which converges. At \( x = 1 \), the series becomes \( \sum_{n=1}^{\infty} 1/n \), which is the harmonic series and diverges.
05

Determining Absolute and Conditional Convergence

The series converges absolutely for \( \frac{1}{3} < x < 1 \) because within this interval, the series converges and the terms \( \frac{(3x-2)^n}{n} \) are positive and decreasing in magnitude. It converges conditionally at \( x = \frac{1}{3} \), as this point makes the series zero, which trivially converges.

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Key Concepts

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

Radius of Convergence
The radius of convergence is a fundamental concept when dealing with power series. It tells us the distance from the center of the series, within which the series will converge.
In the given series,\[ \sum_{n=1}^{\infty} \frac{(3x-2)^n}{n}\]we start by simplifying using the Ratio Test. This process gives us\[ |3x-2| < 1. \]The solution to this inequality shows that the series will converge when the transformed variable \(3x - 2\) lies between -1 and 1.
  • Difference from the center \( \frac{2}{3} \): The maximum difference from the center is 1.
  • Solving for \(x\) results in the interval \( \frac{1}{3} < x < 1 \).
The radius of convergence is thus \( \frac{1}{3} \), highlighting how far from the center \(x = \frac{2}{3}\) the series terms converge absolutely.
Interval of Convergence
Identifying when a series converges requires understanding the interval of convergence. This interval includes all \(x\) values for which the series is convergent and not necessarily only absolutely convergent.
For \( \sum_{n=1}^{\infty} \frac{(3x-2)^n}{n} \), we found through an inequality that the series converges for:
  • \( \frac{1}{3} < x < 1 \)
This gives us its primary interval of convergence.
We must check what happens at the end points of this interval:
  • At \( x = \frac{1}{3} \), the function \[ \sum_{n=1}^{\infty} \frac{0^n}{n} = 0,\]makes it converge trivially.
  • At \( x = 1 \), the function becomes \[ \sum_{n=1}^{\infty} \frac{1}{n},\]which diverges as it represents the harmonic series.
This means the complete interval of convergence is \( \frac{1}{3} \leq x < 1 \). This discovery helps us see clearly where the series behaves differently within and at the bounds of this interval.
Absolute Convergence
Absolute convergence is when the series of absolute values converges. It's a strong form of convergence that ensures stability of the series down its interval.
For our given series, the sequence:\[ \left| \sum_{n=1}^{\infty} \frac{(3x-2)^n}{n} \right| \]can be checked for absolute convergence on \( \frac{1}{3} < x < 1 \).
In this interval, \( |3x-2|<1\) means terms in the series decrease effectively.
  • The convergence behavior is represented by decreasing simplifiable terms, ensuring they sum to a finite limit. Hence, the series converges absolutely.
Absolute convergence here implies the series is robust within this interval and behaves consistently when adding terms without affecting its outcome adversely.
Conditional Convergence
Conditional convergence occurs when a series converges, but not absolutely. This means that the series is converging based on some particular conditions being met.
In our problem,
  • At \( x = \frac{1}{3} \): the series \[ \sum_{n=1}^{\infty} \frac{0^n}{n} = 0 \]certainly converges, because its value is zero. Statistically, any summation to zero converges.
  • However, this cannot be said to be absolute since another transformation here (ignoring zero) would not be equivalent.
Thus, this point is a unique case creating convergence through this rather specific condition, hence reflecting conditional convergence.

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