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Magazine Chapter 9: Problem 27
(a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}}$$
Short Answer
Expert verified
Radius: 2. Interval:
(-4, 0].
Converges absolutely for
-4 < x < 0.
Conditionally at x=0.
Step by step solution
01
Identify the General Term of the Series
The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} \). The general term \( a_n \) of this series can be expressed as \( a_n = \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} \).
02
Use the Ratio Test for Convergence
To find the radius of convergence, apply the Ratio Test. Compute \( \left|\frac{a_{n+1}}{a_n}\right| \):\[\left| \frac{ \frac{(-1)^{n+2}(x+2)^{n+1}}{(n+1)2^{n+1}} }{ \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} } \right| = \left| \frac{(x+2)(n)}{2(n+1)} \right| = \frac{|x+2|}{2}\cdot\frac{n}{n+1}.\] As \( n \to \infty \), \( \frac{n}{n+1} \to 1 \). Thus, the limit is \( \frac{|x+2|}{2} \).
03
Set the Ratio Less Than 1
For the series to converge, we need:\[ \frac{|x+2|}{2} < 1. \]Multiply both sides by 2 to get:\[ |x+2| < 2. \]
04
Solve the Inequality for x
Solve \( |x+2| < 2 \) to find the interval of convergence:\[ -2 < x+2 < 2. \]Subtract 2 from each part:\[ -4 < x < 0. \]
05
Determine the Radius of Convergence
The radius of convergence \( R \) is half the length of the interval of convergence:\( R = \frac{2 - (-2)}{2} = 2 \).
06
Check Endpoints for Absolute Convergence
Check \( x = -4 \) and \( x = 0 \):- **At \( x = -4 \)**: The series becomes \( \sum \frac{(-1)^{n+1}(-2)^n}{n 2^n} \), which is \( \sum \frac{1}{n} \) and diverges.- **At \( x = 0 \)**: The series becomes \( \sum \frac{(-1)^{n+1}2^n}{n 2^n} = \sum \frac{(-1)^{n+1}}{n} \), converging by the Alternating Series Test, so it converges conditionally at \( x=0 \).
07
Evaluate Absolute and Conditional Convergence
- For absolute convergence, check if \( \sum \left| \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} \right| = \sum \frac{|x+2|^n}{n 2^n} \) converges.- For \( x = -4 \), \( \sum \frac{2^n}{n 2^n} = \sum \frac{1}{n} \) diverges, so it doesn't converge absolutely.- For \( |x+2| \geq 2 \), the series diverges; for \( |x+2| < 2 \) it converges absolutely except at the endpoints. At \( x = 0 \), it converges conditionally.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval of convergence
The interval of convergence for a power series is the set of all values of \( x \) for which the series converges. To determine this interval, we often use the Ratio Test. For the series given:
- We first establish the inequality \( \left| \frac{a_{n+1}}{a_n} \right| < 1 \), which involves finding the limit of the ratio of successive terms.
- In this problem, the Ratio Test yields \( \frac{|x+2|}{2} < 1 \).
- Solving this inequality gives us the interval \( -4 < x < 0 \).
Ratio Test
The Ratio Test is a popular method to determine the convergence of series. It's especially useful for series with factorials, powers, or exponential terms. To apply the Ratio Test:
- Calculate the absolute value of the ratio of the \( (n+1) \)-th term to the \( n \)-th term, i.e., \( \left| \frac{a_{n+1}}{a_n} \right| \).
- Simplify to find a limit as \( n \to \infty \).
- If the limit is less than 1, the series converges absolutely.
- If the limit is greater than 1, or infinite, the series diverges.
Absolute convergence
Absolute convergence means that the series \( \sum |a_n| \) converges. In this context, if a series is absolutely convergent, it is also convergent.
- To explore absolute convergence, replace the terms of the original series with their absolute values.
- For this series, we analyze \( \sum \frac{|x+2|^n}{n 2^n} \).
- If \( |x+2| < 2 \), this series converges absolutely.
Conditional convergence
Conditional convergence happens when a series converges, but it does not converge absolutely. This occurs when the series involves alternating signs like in our exercise.
- For the given series, applying the Alternating Series Test at \( x = 0 \) confirms convergence.
- Although the series diverges absolutely at \( x = 0 \), it converges "conditionally" due to alternating terms.
- Conditional convergence indicates that the series converges solely due to its alternating nature, rather than the magnitude of terms.
