Problem 30 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=2}^{\infty} \frac{x^{n}}{n \ln n} $$

Step-by-Step Solution

Answer

The series converges absolutely for \(|x| < 1\), conditionally at \(x = -1\).

Step 1: Identify the General Term

The general term, \(a_n\), of the series \(\sum_{n=2}^{\infty} \frac{x^{n}}{n \ln n}\) is given by \(a_n = \frac{x^n}{n \ln n}\). We will use this to find the series' radius of convergence.

Step 2: Use the Ratio Test

To find the radius of convergence, use the Ratio Test. Calculate the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1) \ln(n+1)} \cdot \frac{n \ln n}{x^n} \right|. \] Simplifying, this becomes: \[ \lim_{n \to \infty} \left| x \cdot \frac{n \ln n}{(n+1) \ln(n+1)} \right|. \] As \(n\) approaches infinity, the expression \(\frac{n \ln n}{(n+1) \ln(n+1)}\) tends to 1. Hence, the limit is \(|x|\). The Ratio Test states the series converges if \(|x| < 1\). Thus, the radius of convergence is 1.

Step 3: Test the Endpoints

Check convergence at the endpoints \(x = -1\) and \(x = 1\). For \(x = -1\), the series becomes \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n}\), which is an alternating series that converges by the Alternating Series Test since \(\frac{1}{n \ln n}\) decreases to 0.For \(x = 1\), the series becomes \(\sum_{n=2}^{\infty} \frac{1}{n \ln n}\), which diverges by the Integral Test, as the integral \(\int_{2}^{\infty} \frac{1}{x \ln x} dx\) diverges.

Step 4: Determine Absolute and Conditional Convergence

Absolute convergence means the series \(\sum |a_n|\) converges. For \(|x| < 1\), the series \(\sum_{n=2}^{\infty} \frac{|x|^n}{n \ln n}\) absolutely converges.At \(x = -1\), the original series converges, but the series \(\sum_{n=2}^{\infty} \frac{1}{n \ln n}\) diverges, indicating conditional convergence.Therefore, for \(|x| < 1\), it converges absolutely, and for \(x = -1\) it converges conditionally.

Key Concepts

Radius of Convergence

The radius of convergence of a power series is crucial to determine the values of \(x\) for which the series will converge. In our exercise, the series is represented by \( \sum_{n=2}^{\infty} \frac{x^{n}}{n \ln n} \). To find the radius of convergence, we start with the general term \( a_n = \frac{x^n}{n \ln n} \). By using the Ratio Test, we calculate the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This simplifies to evaluating the limit \( \lim_{n \to \infty} \left| x \cdot \frac{n \ln n}{(n+1) \ln(n+1)} \right| \), which tends to \(|x|\) as \( n \to \infty \). The Ratio Test tells us the series converges when this limit is less than 1, leading us to the radius of convergence being 1.

Ratio Test

The Ratio Test is a method used to determine the convergence of infinite series. It is particularly useful for series with variable terms like power series. In the given exercise, once the general term \( a_n = \frac{x^n}{n \ln n} \) is identified, the Ratio Test involves computing the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This was simplified to the expression \( \left| x \cdot \frac{n \ln n}{(n+1) \ln(n+1)} \right| \). As \( n \) grows larger, this ratio simplifies to \( |x| \). The series converges absolutely when \( |x| < 1 \), aligning with the result of the Radius of Convergence. Thus, the Ratio Test both guides us to find the radius and serves as a straightforward tool for assessing convergence behavior.

Alternating Series Test

The Alternating Series Test is applied to series whose terms alternate in sign, such as \( \sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n} \) for \( x = -1 \). This test focuses on two main criteria to conclude convergence:

• The absolute value of the terms \( \frac{1}{n \ln n} \) is decreasing.
• The limit of the absolute value of the terms as \( n \to \infty \) approaches zero.

Since both conditions hold, the series for \( x = -1 \) converges. This gives us an example of conditional convergence since the series' positive term version \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) diverges.

Integral Test

The Integral Test helps assess the convergence of series with positive terms. It works best when the terms can be shown as a continuous, positive, and decreasing function. For the exercise, when \( x = 1 \), the series transforms to \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \). By comparing this to the integral \( \int_{2}^{\infty} \frac{1}{x \ln x} \, dx \), we see that it particularly emphasizes that if the integral diverges, so does the series. Therefore, this integral indeed diverges, as it's similar to \( \ln(\ln(x)) \), which grows without bound as \( x \to \infty \). Through the Integral Test, we confirm that the series diverges when \( x = 1 \), adding a layer of understanding to endpoint behavior.