Problem 21 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 2}^{\infty} \frac {1}{n \ln n} \)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify the Series Type

The given series is \( \sum_{n = 2}^{\infty} \frac{1}{n \ln n} \). This series is a form of a harmonic series modified by the natural logarithm.

Step 2: Recognize Test for Convergence

We will use the Integral Test to determine convergence. The Integral Test involves integrating the function \( f(x) = \frac{1}{x \ln x} \), which behaves like the terms of the series, from 2 to infinity.

Step 3: Set Up the Integral

The integral to evaluate is \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \). This will help us determine if the series converges or diverges.

Step 4: Perform Integration

To evaluate \( \int \frac{1}{x \ln x} \, dx \), use substitution: let \( u = \ln x \), thus \( du = \frac{1}{x} \, dx \). The integral becomes \( \int \frac{1}{u} \, du \), which equals \( \ln |u| = \ln |\ln x| \).

Step 5: Evaluate the Improper Integral

Evaluate \( \lim_{b \to \infty} \left( \ln |\ln x| \right) \bigg|_{2}^{b} \). We focus on the behavior as \( x \to \infty \). Substituting, we find \( \ln(\ln(\infty)) - \ln(\ln(2)) \), which approaches infinity.

Step 6: Conclusion from Integral Test

Since the improper integral diverges (it approaches infinity), by the Integral Test, the series \( \sum_{n = 2}^{\infty} \frac{1}{n \ln n} \) diverges.

Key Concepts

Integral Test

The Integral Test is a powerful tool used to determine the convergence or divergence of infinite series. It applies to series whose terms are derived from a continuous, positive, and decreasing function. In our provided solution, the function identified is \( f(x) = \frac{1}{x \ln x} \). To use the Integral Test:

• First, ensure that the function \( f(x) \) matches the sequence of the series terms and is positive and decreasing for all \( x \geq 2 \).
• Next, evaluate the integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \).

If the integral converges (results in a finite number), the series also converges. Conversely, if the integral diverges (approaches infinity), the series also diverges. In our example, the integration showed that the integral diverges. Thus, by the Integral Test, the series \( \sum_{n = 2}^{\infty} \frac{1}{n \ln n} \) also diverges. The test convincingly relates the behavior of the series to the behavior of the integral, providing a clear solution.

Harmonic Series

A harmonic series is one of the fundamental examples of a divergent series. The original harmonic series is expressed as \( \sum_{n=1}^{\infty} \frac{1}{n} \).

Although each term decreases and seems to approach zero, the series itself does not converge to a finite number— it diverges. Now, in the context of the exercise, the series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) can be viewed as a harmonic series modified by a natural logarithm term in the denominator.

• This modification impacts the rate of decrease for each term, affected by the natural logarithmic function, \( \ln n \).
• Despite this extra factor, the series still retains a form that needs rigorous testing for convergence, leading us to apply the Integral Test.

Understanding harmonic series is crucial as it often provides insight into the behavior of similar series that slightly deviate from the harmonic form, emphasizing the importance of tests like the Integral Test to determine drastic outcomes like convergence or divergence.

Natural Logarithm

The natural logarithm, denoted \( \ln(x) \), is a logarithm with Euler's number \( e \approx 2.71828 \) as the base. It's an essential mathematical function that appears often in continuous growth processes, integrals, and complex series. In the context of the given series \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \), the natural logarithm significantly affects the convergence behavior of the series terms.

Here are its critical roles:

• The expression \( \ln n \) in the denominator slows down the decrease of the terms \( \frac{1}{n \ln n} \) compared to the simple harmonic series \( \frac{1}{n} \).
• While \( n \) increases, \( \ln n \) increases more slowly, allowing the series terms to remain relatively larger, relative to the simple harmonic series.

These properties make analyzing the series more complex, necessitating tests like the Integral Test to properly assess convergence. The natural logarithm's behavior highlights the delicate balance of term size that influences whether an infinite series sums up to a finite value or not.