Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=2}^{\infty}(-1)^{n} \frac{1}{n \ln n} $$

Step-by-Step Solution

Answer

The series is conditionally convergent.

Step 1: Define Absolute Convergence

The series \( \sum_{n=2}^{\infty} a_n \) is said to be absolutely convergent if the series of absolute values \( \sum_{n=2}^{\infty} |a_n| \) is convergent. In this case, \( a_n = (-1)^n \frac{1}{n \ln n} \). We need to test the convergence of \( \sum_{n=2}^{\infty} \left| (-1)^n \frac{1}{n \ln n} \right| = \sum_{n=2}^{\infty} \frac{1}{n \ln n} \).

Step 2: Test for Convergence of \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \)

We will test the convergence of the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) using the integral test. Consider the integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \). Let \( u = \ln x \), so \( du = \frac{1}{x} \, dx \), and the integral becomes \( \int \frac{1}{u} \, du \). This integral equals \( \ln |u| + C = \ln (\ln x) + C \). Evaluating from 2 to infinity, the integral diverges, indicating that \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) diverges.

Step 3: Define Conditional Convergence

A series \( \sum_{n=2}^{\infty} a_n \) is conditionally convergent if the series itself converges but the series of its absolute values does not converge. We now need to determine if \( \sum_{n=2}^{\infty} (-1)^n \frac{1}{n \ln n} \) converges.

Step 4: Test for Conditional Convergence Using Alternating Series Test

The alternating series test requires that \( b_n = \frac{1}{n \ln n} \) is decreasing and \( \lim_{n \to \infty} b_n = 0 \). As \( n \) increases, \( n \ln n \) increases, thus \( b_n \) decreases. Also, \( \lim_{n \to \infty} \frac{1}{n \ln n} = 0 \). Since both conditions are satisfied, the series \( \sum_{n=2}^{\infty} (-1)^n \frac{1}{n \ln n} \) converges.

Key Concepts

Absolute Convergence

Absolute convergence of a series occurs when the series formed by taking the absolute values of its terms also converges. Imagine you have a series like \( \sum_{n=1}^{\infty} a_n \) and its related series of absolute values is \( \sum_{n=1}^{\infty} |a_n| \). If this absolute value series converges, we say that the original series is absolutely convergent.

• In simpler terms, if a series is absolutely convergent, then it means no matter the sign (positive or negative) of its terms, the sum will converge.
• Mathematically: A series \( \sum (-1)^n \frac{1}{n \ln n} \) will be absolutely convergent if \( \sum \frac{1}{n \ln n} \) converges.

In our exercise, the series \( \sum \frac{1}{n \ln n} \) was found to diverge using the integral test, so the original series \( \sum (-1)^n \frac{1}{n \ln n} \) is not absolutely convergent.

Conditional Convergence

Conditional convergence is a neat concept where a series converges, but its series of absolute values does not converge. Simply put, the series converges only because of the specific arrangement and sign of its terms.

• Consider our series, \( \sum (-1)^n \frac{1}{n \ln n} \). We already know from testing that \( \sum \frac{1}{n \ln n} \) diverges.
• However, when you assess the series with its alternating sign, it converges. This is what we call conditional convergence.

The inclusion of alternating signs allows the series to eventually settle to a finite sum even when the series of absolute values does not.

Integral Test

The integral test is a powerful method used to determine whether a series converges or diverges using comparison with an improper integral. To use this test, you must have a series \( \sum a_n \) where terms \( a_n = f(n) \) for some positive, continuous, and decreasing function \( f(x) \).

• First, you convert your series into a continuous function \( f(x) \). In our case, \( f(x) = \frac{1}{x \ln x} \).
• Next, evaluate the improper integral \( \int_{2}^{\infty} f(x) \, dx \).
• If this integral converges, then the series \( \sum f(n) \) converges. Conversely, if the integral diverges, so does the series \( \sum f(n) \).

For our series, the integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \) diverges, which aligned perfectly with the outcome showing the series of absolute values also diverges.

Alternating Series Test

The alternating series test helps us find out whether certain alternating series converge. An alternating series is one where the terms keep switching signs, such as \( (-1)^n a_n \).

• For the alternating series \( \sum (-1)^n \frac{1}{n \ln n} \), we have \( a_n = \frac{1}{n \ln n} \).
• The alternating series test requires that \( a_n \) is decreasing, and \( \lim_{n \to \infty} a_n = 0 \).
• In other words, the terms must shrink in a specific order, and approach zero.

In the case of the series given, both conditions are satisfied as \( n \ln n \) increases meaning \( \frac{1}{n \ln n} \) decreases, and \( \lim_{n \to \infty} \frac{1}{n \ln n} = 0 \). Thus, using this test, the series converges due to its alternating nature, even if the absolute series does not.