Problem 6 - Thomas Calculus

Question

Use the Integral Test to determine if the series in Exercises \(1-10\) converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}} $$

Step-by-Step Solution

Answer

The series diverges since the integral diverges to infinity.

Step 1: Check Criteria for Integral Test

For the Integral Test to be applicable, the function that represents the terms of the series must be positive, continuous, and decreasing for all \( n \geq 2 \). In this case, we consider \( f(x) = \frac{1}{x(\ln x)^{2}} \).1. **Positive**: For \( x \geq 2 \), both \( x > 0 \) and \( \ln x > 0 \), so \( (\ln x)^2 > 0 \). Thus, \( f(x) > 0 \) for \( x \geq 2 \).2. **Continuous**: \( f(x) \) is continuous for \( x \geq 2 \) since both the numerator and denominator are continuous on this interval.3. **Decreasing**: To check if \( f(x) \) is decreasing, we find its derivative, \( f'(x) = \frac{d}{dx}\left( \frac{1}{x(\ln x)^{2}} \right) \). Instead, we can ascertain quickly for this function, at least over large values (by examining behavior or testing derivatives), that it indeed decreases.

Step 2: Set Up the Integral

We apply the Integral Test by considering the integral of \( f(x) \) from 2 to infinity. The integral we need to solve is: \[ \int_{2}^{\infty} \frac{1}{x(\ln x)^{2}} \, dx \] If this integral converges, the original series converges. Otherwise, the series diverges.

Step 3: Find an Antiderivative

To evaluate the integral \( \int \frac{1}{x(\ln x)^{2}} \, dx \), let's use substitution. Let \( u = \ln x \), which gives \( du = \frac{1}{x} \, dx \). The integral becomes: \[ \int \frac{1}{u^2} \, du = \int u^{-2} \, du \] The antiderivative of \( u^{-2} \) is \( -u^{-1} + C \), or equivalently, \( -\frac{1}{u} + C \).

Key Concepts

Series Convergence

Understanding series convergence is essential when dealing with infinite series in mathematics. The Integral Test is a common method to determine whether a series converges or diverges. This test applies to series where the terms can be represented by a function that is positive, continuous, and decreasing. In our example, the series is \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}} \).

To use the Integral Test, we express the series terms as \( f(x) = \frac{1}{x(\ln x)^{2}} \) and verify that this function meets the criteria:

• Positive: For all \( x \geq 2 \), both \( x > 0 \) and \( (\ln x)^2 > 0 \), which keeps \( f(x) > 0 \).
• Continuous: The function is continuous for \( x \geq 2 \) since its parts are continuous in this range.
• Decreasing: By analyzing the function or its derivative, it is confirmed that \( f(x) \) decreases.

These conditions ensure we can apply the Integral Test to the series. Calculus tools help us safely move forward to evaluate convergence.

Antiderivative

Finding an antiderivative is a step in solving definite integrals, which in turn helps in checking the convergence of series through the Integral Test. An antiderivative of a function \( f(x) \) is a function whose derivative is \( f(x) \). It’s sometimes referred to as the indefinite integral. In our example, the challenge lies in solving the integral:

\[ \int \frac{1}{x(\ln x)^{2}} \, dx \]

To compute this, we utilize substitution, a common integration method. If we set \( u = \ln x \), then the differential becomes \( du = \frac{1}{x} \, dx \). This transformation simplifies our integral to:

\[ \int \frac{1}{u^2} \, du \]

The antiderivative of \( u^{-2} \) is \( -u^{-1} + C \), or simplified, \( -\frac{1}{u} + C \). This transforms back to \( -\frac{1}{\ln x} + C \) when returning to the original variable. This solution forms a critical step towards evaluating the integral’s convergence.

Calculus Integration

Calculus integration, especially improper integration, plays a crucial role in determining the convergence of series via the Integral Test. Our approach begins with setting up an integral bound from 2 to infinity:

\[ \int_{2}^{\infty} \frac{1}{x(\ln x)^{2}} \, dx \]

This setup requires evaluating an improper integral as the upper limit approaches infinity. Using the techniques of substitution, we transformed and evaluated it in terms of \( u \), leading to

• \[ \lim_{b \to \infty} \left[ -\frac{1}{\ln x} \right]_{2}^{b} \]
• \[ \lim_{b \to \infty} \left( -\frac{1}{\ln b} + \frac{1}{\ln 2} \right) \]

Observing limits is crucial for such integrals. As \( b \to \infty \), \( -\frac{1}{\ln b} \to 0 \). Thus, the integral converges, signifying that the series converges as well. Integration techniques are foundational for judiciously evaluating infinite series' convergence via the Integral Test.