Problem 30 - Calculus: Early Transcendentals

Question

Find the values of \( p \) for which the series is convergent. \( \displaystyle \sum_{n = 3}^{\infty} \frac {{1}}{{ n \ln n [\ln (\ln n)]^P}} \)

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Series

The series given is \( \sum_{n=3}^{\infty} \frac{1}{n \ln n [\ln(\ln n)]^P} \). This looks like a generalized form of a \( p \)-series. The key is identifying the role of the exponent \( P \) on the function \( [\ln(\ln n)]^P \).

Step 2: Apply Integral Test

The integral test states that for a series \( \sum a_n \), if \( a_n = f(n) \) where \( f(x) \) is a continuous, positive, decreasing function, then \( \sum a_n \) converges if and only if the integral \( \int f(x) \, dx \) converges. Here, use \( f(x) = \frac{1}{x \ln x [\ln(\ln x)]^P} \).

Step 3: Set Up Integral

Consider the integral from 3 to infinity: \( \int_{3}^{\infty} \frac{1}{x \ln x [\ln(\ln x)]^P} \, dx \). Rewriting the integral helps us determine the convergence by focusing on its behavior as \( x \to \infty \).

Step 4: Analyze the Integral

Change of variable can be beneficial. Let \( u = \ln(\ln x) \), so \( du = \frac{1}{\ln x} \cdot \frac{1}{x} \, dx \). The integral becomes \( \int \frac{1}{u^P} \, du \).

Step 5: Determine Convergence

The integral \( \int \frac{1}{u^P} \, du \) converges when \( P > 1 \), since it is a \( p \)-integral where convergence occurs for \( P > 1 \) and diverges for \( P \leq 1 \).

Step 6: Conclusion on Series Convergence

Since the integral converges when \( P > 1 \), the given series also converges for \( P > 1 \). If \( P \leq 1 \), the integral diverges; thus the series diverges as well.

Key Concepts

integral test

The integral test is a fundamental concept used in calculus to determine the convergence of infinite series. Specifically, this test is applicable to a series of the form \( \sum a_n \) when you can express \( a_n \) as a function \( f(n) \), where \( f(x) \) is continuous, positive, and decreasing. By comparing the series to the improper integral \( \int f(x) \, dx \), you conclude:

• If the integral \( \int f(x) \, dx \) converges, then the series \( \sum a_n \) also converges.
• If the integral \( \int f(x) \, dx \) diverges, then the series \( \sum a_n \) diverges as well.

In this context, apply the integral test to \( f(x) = \frac{1}{x \ln x [\ln(\ln x)]^P} \). By examining the behavior of the integral from 3 to infinity, it reveals whether the original series converges or diverges. This application is crucial as it provides a bridge between discrete sums and continuous integration, highlighting the behavior of series parallel to functions.

p-series

A \( p \)-series, denoted as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), is a specific type of series dependent on the parameter \( p \). Recognizing a \( p \)-series is significant, as the convergence depends solely on the value of \( p \). The primary guidelines are:

• If \( p > 1 \), the \( p \)-series converges.
• If \( p \leq 1 \), the \( p \)-series diverges.

In the exercise discussed, although the series formula is different, it similarly reduces to a \( p \)-series through variable substitution, bringing to light the role of \( P \) in the convergence criteria. Recognizing this concept allows us to anchor our understanding of series behavior to the simple guidelines of a \( p \)-series.

change of variable

The change of variable technique is a powerful mathematical tool used to simplify integrals and make them more tractable. In this scenario, we apply the substitution method by letting \( u = \ln(\ln x) \), transforming the original integral \( \int_{3}^{\infty} \frac{1}{x \ln x [\ln(\ln x)]^P} \, dx \). This leads to the simplified integral \( \int \frac{1}{u^P} \, du \).
This technique is particularly useful as it shifts the problem from handling complex expressions to dealing with more straightforward forms.

• Reduces complex expressions into recognizable forms, aiding in solving.
• Simplifies the evaluation of integrals, especially when aligning them with known convergence behaviors.

Using substitutions correctly will prepare us for identifying convergence by transforming a challenging problem into an understandable one through the lens of familiar mathematical constructs.

convergence criterion

Understanding the convergence criterion is essential in determining whether a series or an integral converges or diverges. The convergence of an integral is determined based on the behavior of the integrand and the limits of integration. For the given integral \( \int \frac{1}{u^P} \, du \), it converges if \( P > 1 \) and diverges for \( P \leq 1 \). This aligns with the knowledge from \( p \)-series, but here applied to integrals directly.

• Relies on known mathematical formulas and tests.
• Enables prediction of series behavior with precise conditions.

By recognizing these criteria, we can make informed decisions about the nature of mathematical series and integrals. Understanding the criteria guides our resolution of problems, leading to correct conclusions about their convergence or divergence.