Problem 8
Question
In Exercises \(1-36,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely (c) conditionally? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}(x+2)^{n}}{n} $$
Step-by-Step Solution
Verified Answer
The series converges absolutely for \((-3, -1]\). It does not converge conditionally.
1Step 1: Understanding the series
The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(x+2)^{n}}{n} \). This is an alternating series, given that the terms have a factor of \((-1)^n\). The general form is \( a_n = \frac{(-1)^n (x+2)^n}{n} \).
2Step 2: Finding the radius of convergence using the Ratio Test
The Ratio Test states that a series \( \sum a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). For the series, we calculate:\[ \frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} (x+2)^{n+1} / (n+1)}{(-1)^n (x+2)^n / n} = \frac{(x+2)}{1} \cdot \frac{n}{n+1} \]Therefore,\[ \left| \frac{a_{n+1}}{a_n} \right| = |x+2| \cdot \frac{n}{n+1} \]Taking the limit as \( n \to \infty \), we get:\[ |x+2| \cdot \lim_{n \to \infty} \frac{n}{n+1} = |x+2| \]For convergence, \( |x+2| < 1 \).
3Step 3: Determine the interval of convergence
From \(|x+2| < 1\), solve for \(x\):\[ -1 < x+2 < 1 \]Subtracting 2 throughout gives:\[ -3 < x < -1 \]Therefore, the interval of convergence is \((-3, -1)\).
4Step 4: Test endpoints for absolute and conditional convergence
To determine conditional and absolute convergence, test \(x = -3\) and \(x = -1\).1. **At \(x = -3\):** The series becomes \[ \sum_{n=1}^{\infty} \frac{(-1)^{n}(-1)^{n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \] This is the Harmonic series, which diverges.2. **At \(x = -1\):** The series becomes \[ \sum_{n=1}^{\infty} \frac{(-1)^{n}0^{n}}{n} \] Since terms are zero for all \(n\), the series converges absolutely.
5Step 5: Conclude values of x for absolute and conditional convergence
- The series converges absolutely for \(-3 < x < -1\) and \(x = -1\) since all terms are zero.- The series does not converge at \(x = -3\), hence conditional convergence is not applicable.
Key Concepts
Alternating SeriesRatio TestAbsolute ConvergenceConditional Convergence
Alternating Series
Let's start by understanding what an alternating series is. An alternating series is a series whose terms alternate in sign. The typical form of an alternating series is \( a_n = (-1)^n b_n \) or \((-1)^{n+1} b_n\), where \(b_n\) are positive terms. These series can have fascinating properties because the alternating nature affects convergence. The series given in the exercise \( \sum_{n=1}^{\infty} \frac{(-1)^n (x+2)^n}{n} \) is indeed an alternating series, as you can see the presence of the \((-1)^n\) factor.
Alternating series are particularly interesting because they can converge even if the non-alternating version diverges. This makes them special when determining convergence using specific tests like the Alternating Series Test.
Alternating series are particularly interesting because they can converge even if the non-alternating version diverges. This makes them special when determining convergence using specific tests like the Alternating Series Test.
Ratio Test
The Ratio Test is a powerful tool to determine the absolute convergence of a series. It evaluates the limit of the ratio of consecutive terms of the series. The test states that for \( \sum a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges absolutely. Conversely, if this limit is greater than one, the series diverges. If it equals one, the test is inconclusive.
For our series, by using the Ratio Test, we calculated:\( \left| \frac{a_{n+1}}{a_n} \right| = |x+2| \cdot \frac{n}{n+1} \). As \(n\) approaches infinity, \(\frac{n}{n+1}\) approaches 1, leaving us with \(|x+2|\). The condition for the series to converge is \(|x+2| < 1\), leading us to the interval of convergence.
For our series, by using the Ratio Test, we calculated:\( \left| \frac{a_{n+1}}{a_n} \right| = |x+2| \cdot \frac{n}{n+1} \). As \(n\) approaches infinity, \(\frac{n}{n+1}\) approaches 1, leaving us with \(|x+2|\). The condition for the series to converge is \(|x+2| < 1\), leading us to the interval of convergence.
Absolute Convergence
Absolute convergence takes a deeper look into how series behave. If a series converges absolutely, it means the series of absolute values \(\sum |a_n|\) converges. This is a stronger form of convergence and has nice properties, such as preserving convergence under rearrangement of terms.
For the given series \( \sum_{n=1}^{\infty} \frac{(-1)^n (x+2)^n}{n} \), absolute convergence was tested using endpoints. At \(x = -1\), all terms become 0, making the series absolutely convergent. The series also converges absolutely for \(-3 < x < -1\) due to the outcome from the Ratio Test. Absolute convergence gives assurance because it indicates the series sums up neatly, regardless of term order.
For the given series \( \sum_{n=1}^{\infty} \frac{(-1)^n (x+2)^n}{n} \), absolute convergence was tested using endpoints. At \(x = -1\), all terms become 0, making the series absolutely convergent. The series also converges absolutely for \(-3 < x < -1\) due to the outcome from the Ratio Test. Absolute convergence gives assurance because it indicates the series sums up neatly, regardless of term order.
Conditional Convergence
Conditional convergence occurs when a series converges, but its absolute counterpart does not. This kind of series is sensitive to changes like term order rearrangement. In our exercise, we needed to check for endpoints where only conditional convergence may occur.
We tested \(x = -3\) by transforming the series into the well-known Harmonic series: \(\sum \frac{1}{n}\), which diverges. Hence, at this endpoint, neither absolute nor conditional convergence was present. Understanding conditional convergence is crucial in real analysis and demonstrates that convergence is not just about summing up numbers, but how they are combined.
We tested \(x = -3\) by transforming the series into the well-known Harmonic series: \(\sum \frac{1}{n}\), which diverges. Hence, at this endpoint, neither absolute nor conditional convergence was present. Understanding conditional convergence is crucial in real analysis and demonstrates that convergence is not just about summing up numbers, but how they are combined.
Other exercises in this chapter
Problem 8
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