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Chapter 9: Problem 22

(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 \( R = \frac{1}{9} \), interval \( \left( \frac{17}{9}, \frac{19}{9} \right) \). Absolute convergence for \( x \neq \frac{17}{9}, \frac{19}{9} \), conditional at endpoints.

Step by step solution

01

Identify the Form of the Series

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n} 3^{2n}(x-2)^{n}}{3n} \). This is a power series centered at \( x=2 \). The general form is \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where \( c=2 \) in this case.
02

Apply the Ratio Test for Absolute Convergence

The ratio test involves finding the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_n = \frac{(-1)^n 3^{2n} (x-2)^n}{3n} \). Calculate \( a_{n+1} = \frac{(-1)^{n+1} 3^{2(n+1)} (x-2)^{n+1}}{3(n+1)} \).
03

Compute the Ratio

Substitute \( a_{n} \) and \( a_{n+1} \) into the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3^{2(n+1)} (x-2)^{n+1} \cdot 3n}{3^{2n} (x-2)^{n} \cdot 3(n+1)} \right| = \left| \frac{9(x-2) \cdot n}{n+1} \right| \].
04

Simplify the Ratio

As \( n \to \infty \), \( \frac{n}{n+1} \approx 1 \). Therefore, the limit simplifies to \( 9|x-2| \). Set \( 9|x-2| < 1 \) to find the interval of convergence.
05

Solve for the Radius and Interval of Convergence

Solving \( 9|x-2| < 1 \) gives \( |x-2| < \frac{1}{9} \). This means the radius \( R = \frac{1}{9} \), and the interval of convergence is \( 2-\frac{1}{9} < x < 2+\frac{1}{9} \) or \( \frac{17}{9} < x < \frac{19}{9} \).
06

Test Endpoints for Conditional and Absolute Convergence

Test \( x = \frac{17}{9} \) and \( x = \frac{19}{9} \). Substitute these values and check if the series converges absolutely (converges regardless of the sign) or conditionally (only converges when alternating signs are considered).
07

Conclusion on the Type of Convergence

At the endpoints, \( x = \frac{17}{9} \) and \( x = \frac{19}{9} \), substitute back into the series and check convergence. Generally, if the absolute value of the series converges at these points, it's absolute convergence; otherwise, check for conditional convergence using the alternating series test.

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Key Concepts

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

Ratio Test
The Ratio Test is an important tool for determining the convergence of series, especially when dealing with power series or series with factorial numbers and exponential functions. Here's how it works:

  • To apply the Ratio Test, look at the limit of the absolute value of the ratio of consecutive terms in the series: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
  • If \( L < 1 \), the series converges absolutely. If \( L > 1 \) or \( L = \infty \), the series diverges.
  • If \( L = 1 \), the Ratio Test is inconclusive, and we may need other tests to determine convergence.
In the original example, the Ratio Test was applied to \( \sum_{n=1}^{\infty} \frac{(-1)^{n} 3^{2 n}(x-2)^{n}}{3 n} \). We found that simplifying the limit gave us \( L = 9|x-2| \). This helped to find the convergence properties of the series.
Interval of Convergence
The interval of convergence relates to power series' limits within which the series converges. Let's break down the process of finding it.
  • Using the Radius of Convergence obtained from the Ratio Test, sprinkle it around the center of the series. This series was centered at \(x = 2\).
  • The inequality \(9|x-2| < 1\) was solved to determine the interval of convergence, resulting in \(\frac{17}{9} < x < \frac{19}{9}\).
  • Within this interval, the series converges absolutely as confirmed by the Ratio Test.
To be thorough, always test the endpoints of the interval separately to determine convergence behavior.
Conditional Convergence
Conditional convergence is a fascinating concept, referring to when a series converges with alternating terms while not converging absolutely.

  • To check for conditional convergence at the endpoints, consider substituting these into the series.
  • In the provided exercise, for \(x = \frac{17}{9}\) and \(x = \frac{19}{9}\), it鈥檚 essential to observe if the series converges only due to the alternating nature of its terms.
  • The Alternating Series Test is commonly used here: check if the absolute values of terms decrease to zero.
Ensure to differentiate between absolute and conditional convergence for comprehensive understanding. This distinction is pivotal because a series that converges conditionally must meet specific criteria not necessary for absolute convergence.

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Most popular questions from this chapter

What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.

It is not yet known whether the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \sin ^{2} n} $$converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps. a. Define the sequence of partial sums $$ s_{k}=\sum_{n=1}^{k} \frac{1}{n^{3} \sin ^{2} n} $$ What happens when you try to find the limit of \(s_{k}\) as \(k \rightarrow \infty ?\) Does your CAS find a closed form answer for this limit? b. Plot the first 100 points \(\left(k, s_{k}\right)\) for the sequence of partial sums. Do they appear to converge? What would you estimate the limit to be? c. Next plot the first 200 points \(\left(k, s_{k}\right) .\) Discuss the behavior in your own words. d. Plot the first 400 points \(\left(k, s_{k}\right) .\) What happens when \(k=355 ?\) Calculate the number \(355 / 113 .\) Explain from you calculation what happened at \(k=355 .\) For what values of \(k\) would you guess this behavior might occur again?

Show by example that \(\Sigma\left(a_{n} / b_{n}\right)\) may converge to something other than \(A / B\) even when \(A=\Sigma a_{m}, B=\Sigma b_{n} \neq 0,\) and no \(b_{n}\) equals 0

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