Question

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n}\left(1-n^{1 / n}\right)\) (Hint: \(n^{1 / n} \approx 1+\ln (n) / n\) for large \(\left.n .\right)\)

Short answer

The series converges conditionally.

Step-by-step solution

Understanding the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n}(1-n^{1/n}) \). This series is alternating due to the factor \((-1)^n\). We determine convergence by testing if the absolute version of the series is convergent and then test for conditional convergence.

Examine the Term's Behavior

We need to understand the behavior of the term \( 1-n^{1/n} \) as \( n \to \infty \). Using the hint, \( n^{1/n} \approx 1+\frac{\ln(n)}{n} \), so \( 1-n^{1/n} \approx 1-(1+\frac{\ln(n)}{n}) = -\frac{\ln(n)}{n} \). This implies \( 1-n^{1/n} \) behaves approximately like \(-\frac{\ln(n)}{n}\) for large \(n\).

Absolute Convergence Test

Consider the absolute series \( \sum_{n=1}^{\infty} \left| (-1)^{n} (1-n^{1/n}) \right| = \sum_{n=1}^{\infty} \left| 1-n^{1/n} \right| \approx \sum_{n=1}^{\infty} \frac{\ln(n)}{n}\). The comparison test or integral test shows that \( \sum \frac{\ln(n)}{n} \) diverges because \( \int \frac{\ln(x)}{x} \, dx \) diverges.

Check Conditional Convergence

The Alternating Series Test is used next. For the series \( \sum_{n=1}^{\infty}(-1)^{n}u_{n} \) with \( u_n = 1-n^{1/n} \approx -\frac{\ln(n)}{n} \), we check if \( u_n \to 0 \) and if \( |u_n| \) is decreasing for large \( n \). Since \( u_n \propto \frac{\ln(n)}{n} \), as \( n \to \infty \), \( |u_n| \to 0 \), and \( u_n \) decreases. Hence, the series converges conditionally.

Key concepts

Absolute Convergence

When we discuss absolute convergence, we are referring to whether the series remains convergent even if we take the absolute values of all its terms. In mathematical terms, a series \( \sum_{n=1}^{\infty} a_n \) is said to converge absolutely if the series \( \sum_{n=1}^{\infty} |a_n| \) converges. For the given series \( \sum_{n=1}^{\infty}(-1)^{n}(1-n^{1/n}) \), this means evaluating \( \sum_{n=1}^{\infty} \left|1-n^{1/n}\right| \). With the approximation \( n^{1/n} \approx 1+\frac{\ln(n)}{n} \) for large \( n \), it follows that \( 1-n^{1/n} \approx -\frac{\ln(n)}{n} \). This implies that the absolute series behaves like \( \sum_{n=1}^{\infty} \frac{\ln(n)}{n} \). By using convergence tests such as the Comparison Test or the Integral Test, we find that \( \sum \frac{\ln(n)}{n} \) diverges. Therefore, the series does not converge absolutely.

Conditional Convergence

Conditional convergence occurs when a series converges, but does not converge absolutely. It's an important concept when dealing with alternating series. In the case of our series \( \sum_{n=1}^{\infty}(-1)^{n}(1-n^{1/n}) \, \) we want to know if despite failing to converge absolutely, it converges under the right conditions.The Alternating Series Test helps us here. It states that for a series \( \sum_{n=1}^{\infty} (-1)^{n} b_n \, \) if

• the terms \( |b_n| \) are decreasing, and
• \( |b_n| \to 0 \, \)

then the series converges. For our series, with terms \( u_n = 1-n^{1/n} \approx -\frac{\ln(n)}{n} \), as \( n \to \infty \), it's clear \( |u_n| \to 0 \) and \( |u_n| \) is decreasing. Hence, the series meets the criteria for conditional convergence, signifying that it converges when the alternating sign is maintained, even though the absolute series does not.

Alternating Series Test

The Alternating Series Test is crucial when dealing with series that change signs, such as those with terms like \((-1)^n\). This test helps us decide if such a series converges. The test requires two main conditions:

• The sequence \(|b_n| \) must be decreasing: each term in absolute value is smaller than the previous.
• The terms \(|b_n| \to 0\) as \( n \to \infty \): the general term should tend to zero.

For the series \( \sum_{n=1}^{\infty} (-1)^{n} (1-n^{1/n})\), when examining \( u_n = 1-n^{1/n} \approx -\frac{\ln(n)}{n} \), \( |u_n| \) becomes \( \frac{\ln(n)}{n} \. \)As \( n \to \infty \), these terms clearly decrease and approach zero, satisfying both conditions of the Alternating Series Test. Therefore, despite lacking absolute convergence, the series converges conditionally thanks to the nature of its alternating sequence.