Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

For what values of \(p\) does \(\sum_{n=2}^{\infty} 1 /\left[n(\ln n)^{p}\right]\) converge? Explain.

Step-by-Step Solution

Answer

The series converges for \( p > 1 \).

Step 1: Identify the series form and comparison test

The given series is \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} \). We need to determine for which values of \( p \) this series converges. Knowing it's a variety of the p-series or logarithmic variant, we use a comparison test to a known convergent series.

Step 2: Compare with an improper integral

Apply the integral test for convergence. Consider the function \( f(x) = \frac{1}{x(\ln x)^{p}} \), which is positive, continuous, and decreasing for \( x \geq 2 \). Evaluate the improper integral \( \int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} \, dx \).

Step 3: Change of variables in the integral

Use a substitution to simplify the integral. Let \( u = \ln x \), then \( du = \frac{1}{x}dx \). The limits change from \( x=2 \) to \( x=\infty \) becoming \( u=\ln 2 \) to \( u = \infty \). The integral is reformulated as \( \int_{\ln 2}^{\infty} \frac{1}{u^p} \, du \).

Step 4: Evaluate the integral for convergence

The integral \( \int_{\ln 2}^{\infty} \frac{1}{u^p} \, du \) converges if \( p > 1 \) and diverges for \( p \leq 1 \). This follows from the basic p-integral convergence test, where \( \int_{a}^{b} \frac{1}{u^p} \, du \) converges if \( p > 1 \).

Step 5: Conclude the convergence of the series

Based on the evaluation of the integral, the original series \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} \) converges if and only if \( p > 1 \).

Key Concepts

Improper Integral

When working with series, especially to determine convergence, improper integrals are a powerful tool. Consider an integral where either the limits or the function itself approaches infinity. This is what we term an improper integral. For example, the integral \( \int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} \, dx \) is improper because the integration limit extends to infinity. Using improper integrals helps us analyze series. If you can express a series as the sum of function values, comparing it to an integral can help determine convergence:

• If the integral converges, so does the series.
• If the integral diverges, so does the series.

This applies when the function is positive, continuous, and decreasing on the interval considered. For our purposes, using improper integrals with series allows us to translate a difficult summation problem into a potentially simpler integration problem.

Comparison Test

The Comparison Test is a method to determine the convergence of a series by comparing it to another series with known behavior. Essentially, if you've worked with one series and already know whether it converges or diverges, you can use it as a benchmark.

• If \( a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \) for all \( n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In practical terms, the Comparison Test requires choosing an appropriate comparison—one that mirrors the series' difficulty and known outcomes. This is where knowing if a basic series like the harmonic or geometric series converges can become handy. For the given problem, the consideration of \( \frac{1}{x(\ln x)^{p}} \) was compared to a known test integral, allowing the Conversion Test to affirm convergence when \( p > 1 \). This known integral transforms to resemble a p-series, assisting in determining the limits for \( p \).

p-Series

p-Series are specific types of series defined as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where the value of \( p \) determines convergence. When \( p > 1 \), the p-series converges. On the flip side, when \( p \leq 1 \), the series diverges.
This power series is a staple item in calculus due to its straightforward formulation and easy-to-remember criteria for convergence. For series involving logarithmic expressions, it acts as a stepping stone to determine behavior. By comparing a series involving multiple factors, such as both power and log terms, to a simple \( p \)-series, you can often simplify your work. Understanding \( p \)-series and their behavior helps in comparison tactics, especially when transforming or working with more elaborate expressions. In our series, recognizing the role the logarithmic function plays allows us to tie back to these familiar cases.

Substitution Method

The Substitution Method is a technique in calculus used to make integration easier and more straightforward, especially with complex functions or expressions. Here's how it works: You replace a difficult part of the integral with a simpler variable, helping make the integral more manageable. In the context of our problem, by letting \( u = \ln x \) and subsequently finding \( du = \frac{1}{x}dx \), we transformed the complex integral \( \int_2^{\infty} \frac{1}{x (\ln x)^p} \, dx \) to the more straightforward \( \int_{\ln 2}^{\infty} \frac{1}{u^p} \, du \).

• This change simplifies the variable dependencies.
• Shifts the focus to a simpler arithmetic or algebraic concept.

Through substitution, we shift the problem's constraints and integrate with new limits or terms. Once the new integral is evaluated, the original expression's convergence or divergence often becomes clearer.