Question

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1} \ln (1 / n)\)

Short answer

The series converges conditionally.

Step-by-step solution

Identify the Type of Series

The series given is \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln (1 / n)\). This is an alternating series because it has the form \((-1)^{n+1}a_n\) where \(a_n = \ln\left(\frac{1}{n}\right)\).

Check for Absolute Convergence

To check for absolute convergence, consider the absolute value of the series: \(\sum_{n=1}^{\infty} \left|(-1)^{n+1} \ln (1/n)\right| = \sum_{n=1}^{\infty} \ln (1/n)\). This simplifies to \(\sum_{n=1}^{\infty} -\ln(n)\). As \(n\) approaches infinity, \(-\ln(n)\) approaches infinity, thus the series \(\sum_{n=1}^{\infty} -\ln(n)\) diverges.

Apply the Alternating Series Test

To check whether the original series converges conditionally, use the Alternating Series Test. The test requires two conditions: (1) \(a_n = \ln\left(\frac{1}{n}\right)\) is eventually decreasing, and (2) \(\lim_{n\to\infty} a_n = 0\).1. Since \(a_n = \ln\left(\frac{1}{n}\right) = -\ln(n)\), it is decreasing for all \(n \geq 2\).2. \(\lim_{n\rightarrow\infty} a_n = \lim_{n\to\infty} \ln\left(\frac{1}{n}\right) = 0\).Both conditions are satisfied, so the series converges.

Key concepts

Alternating Series Test

An alternating series is a series whose terms alternate in sign. The general form is \((-1)^n a_n\), where \(a_n\) is a sequence of positive terms. To determine the convergence of such a series, we use the Alternating Series Test. The test states that an alternating series converges if two conditions are met:

• The absolute value of terms \(a_n\) is eventually decreasing. This means that each term in the sequence does not grow larger as the series progresses.
• The limit of \(a_n\) as \(n\) approaches infinity is zero: \(\lim_{n \to \infty} a_n = 0\).

If both conditions are satisfied, the series converges. This means it approaches a specific number as \(n\) becomes very large. In our example with \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln(1/n)\), the conditions are met, so the series converges. Despite not converging absolutely, it still holds together by alternating its sign.

Absolute Convergence

Absolute convergence in a series occurs when the series of absolute values also converges. To test for absolute convergence, we take the absolute value of each term and see if the series formed converges. For a series \(\sum_{n=1}^{\infty} a_n\), if the series \(\sum_{n=1}^{\infty} |a_n|\) converges, then the original series converges absolutely. This type of convergence is quite strong and suggests that terms are shrinking rapidly enough for convergence.In the exercise series \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln (1/n)\), we considered the series \(\sum_{n=1}^{\infty} -\ln(n)\) for absolute convergence. Since \(-\ln(n)\) tends towards infinity as \(n\) increases, this series diverges, indicating that the original series does not converge absolutely.

Conditional Convergence

Conditional convergence occurs when a series converges, but it does not converge absolutely. In other words, the series \(\sum_{n=1}^{\infty} a_n\) converges, yet \(\sum_{n=1}^{\infty} |a_n|\) diverges. This type of convergence relies on the specific properties of an alternating series to bring terms towards a limit, despite the lack of absolute shrinkage of term values.In our example of \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln(1/n)\), we found that it passed the Alternating Series Test but failed the absolute convergence test. Thus, it converges conditionally. Conditional convergence indicates a delicate balance; removing the alternating sign would cause the series to diverge.

Divergence

Divergence is a scenario where a series does not converge to a finite value. Instead, it may keep increasing indefinitely, or it simply does not settle towards any particular number. Divergence indicates instability in trying to bring the infinite sum to a conclusion.In the context of absolute convergence, we tested the series \(\sum_{n=1}^{\infty} \ln(1/n)\), which simplified to \(\sum_{n=1}^{\infty} -\ln(n)\). As \(n\) becomes very large, \(-\ln(n)\) approaches infinity. This means that the series diverges. Therefore, while the original series \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln(1/n)\) converges conditionally, its absolute counterpart diverges.

Logarithmic Functions

Logarithmic functions play a pivotal role in analyzing series convergence, particularly when they appear in the formulas for terms of a series. A logarithm, particularly the natural logarithm \(\ln(n)\), grows slowly as \(n\) increases. This slow growth impacts how terms in a series behave in the long term.In the given exercise, \(\ln(1/n)\) simplifies to \(-\ln(n)\). This transformation helps in recognizing convergence patterns. Knowing that \(\ln(n)\)'s growth eventually leads towards infinity is key to understanding why \(\sum_{n=1}^{\infty} \ln(1/n)\) diverges, contributing to our assessment of absolute convergence. Understanding how logarithms behave in series terms helps in predicting and testing for convergence more effectively.