Problem 22
Question
In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n-1}{n} $$
Step-by-Step Solution
Verified Answer
The series is divergent.
1Step 1: Identify the Series Type
The series given is \[\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n-1}{n}\]which is an alternating series. We need to determine if it is absolutely convergent, conditionally convergent, or divergent. The alternating series test is a good starting point.
2Step 2: Check Absolute Convergence
To check for absolute convergence, we need to consider the series without the alternating sign:\[\sum_{n=1}^{\infty} \frac{n-1}{n}\]This simplifies to:\[\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)\]This series does not converge because the main part behaves like \(1 - \frac{1}{n}\) for large \(n\), further, since \(\frac{1}{n}\) diverges, \(\sum \left(1 - \frac{1}{n}\right)\) diverges. Thus, the series is not absolutely convergent.
3Step 3: Apply the Alternating Series Test
Apply the alternating series test:1. Check if \(\frac{n-1}{n}\) is decreasing. As \(n\) increases \(\frac{n-1}{n}\to 1\), the sequence is decreasing because each successive term is slightly smaller as the \(-\frac{1}{n}\) part gets larger.2. Check if \(\lim_{n \to \infty} \frac{n-1}{n} = 0\). The limit actually equals 1.Since the limit does not equal zero, the alternating series test shows that the series diverges.
Key Concepts
absolute convergenceconditional convergencealternating series test
absolute convergence
Absolute convergence is a significant concept in the study of series. When we say that a series is absolutely convergent, we are saying that the series converges even when we take the absolute value of each of its terms. To check for absolute convergence, we eliminate the alternating sign and consider the series of absolute values.
For instance, consider the series:
For instance, consider the series:
- \[egin{equation} ext{Original series: } onumber\ ext{Given } extstyle ext{series is } extstyle extstyle extstyle ext{} extstyle onumber\ ext{We take the absolute value: } onumber\ extstyle extstyle extstyle extstyle extstyle extstyle onumber\ extstyle extstyle onumber\ extstyle ext{Check if this new series converges} onumber\ extstyle extstyle\ \ end{equation}\]
- n is non-integral of 0
- The testing symbolizes pointwise testing for convergence
- Classically continued fractions dictating areas arise
conditional convergence
Conditional convergence occurs when a series converges, but does not converge absolutely. This means that while considering the alternating components, the series may converge; however, without them, it doesn't.
Let's consider the original series again:
The fact that a series is conditionally convergent provides insights into the pivotal role that alternating operations play in leading the series to converge. It is essential for students to comprehend the significance of alternating signs in determining convergence conditions.
Let's consider the original series again:
- For this case, when we remove the alternating factor, we test the series
- aped hoping zealously series,
- also definitions exhibit to the indefinite
- \( extstyle ext{with } ext \)
The fact that a series is conditionally convergent provides insights into the pivotal role that alternating operations play in leading the series to converge. It is essential for students to comprehend the significance of alternating signs in determining convergence conditions.
alternating series test
The alternating series test is a helpful tool to determine the convergence of a series that alternates in sign. This test is particularly applicable when a series alternates between positive and negative terms, as was the case with the original exercise problem.
For the series \((-1)^{n+1} \frac{n-1}{n}\), we evaluate using two main criteria of the alternating series test:
For the series \((-1)^{n+1} \frac{n-1}{n}\), we evaluate using two main criteria of the alternating series test:
- The sequence of terms \(\frac{n-1}{n}\) must be decreasing.
- As \(n\) increases, the value heads towards 1, implying a slight decrease.
- The limit of the terms as \(n\) approaches infinity must be zero.
- However, instead of zero, we find it approaches 1, indicating divergence of the alternating series because of continued non-zero convergence.
Other exercises in this chapter
Problem 22
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