Problem 6 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Direct Comparison Test to determine if each series converges or diverges. $$\sum_{n=1}^{\infty} \frac{1}{n 3^{n}}$$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{n 3^{n}} \) converges by the Direct Comparison Test.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n 3^{n}} \). We aim to determine whether this series converges or diverges using the Direct Comparison Test.

Step 2: Choose a Comparison Series

To use the Direct Comparison Test, we need another series \( \sum_{n=1}^{\infty} b_n \) to compare with \( \sum_{n=1}^{\infty} a_n \), where \( a_n = \frac{1}{n 3^{n}} \). We choose the geometric series \( \sum_{n=1}^{\infty} \frac{1}{3^{n}} \), which can be rewritten as \( \sum_{n=1}^{\infty} b_n \) with \( b_n = \frac{1}{3^{n}} \). This series converges because it is a geometric series with ratio \( r = \frac{1}{3} < 1 \).

Step 3: Compare the Terms

We will compare the terms \( a_n = \frac{1}{n 3^{n}} \) with \( b_n = \frac{1}{3^{n}} \). Notice that since \( n \geq 1 \), \( n \geq 1 \cdot 3^0 = 1 \), hence \( \frac{1}{n 3^{n}} \leq \frac{1}{3^{n}} \).

Step 4: Apply the Direct Comparison Test

Since \( 0 \leq a_n \leq b_n \) for all \( n \geq 1 \) and \( \sum_{n=1}^{\infty} b_n \) converges, by the Direct Comparison Test, \( \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{n 3^{n}} \) also converges.

Key Concepts

Convergence of Series

A series is essentially an infinite sum of terms. When we talk about the convergence of a series, we are asking whether the infinite sum approaches a specific number, known as the limit, as we add more and more terms. If the series has a limit, it converges; otherwise, it diverges.

In solving problems about convergence, we use various tests. One such test is the Direct Comparison Test. By comparing a series to another one with known behavior, we can determine whether our series converges or diverges.

• A convergent series has a finite sum as the number of terms increases.
• A divergent series does not have a finite sum, continuing indefinitely.

Understanding the convergence of a series helps us in various mathematical fields, such as calculus, where we analyze the behavior of functions and sequences.

Geometric Series

A geometric series is a special type of series where each term is a constant multiple of the term before it. This constant is known as the common ratio. A simple geometric series might look like this: \[ 1 + r + r^2 + r^3 + \, \]where \(r\) represents the common ratio.

The behavior of a geometric series mainly depends on the value of \(r\). If \(|r| < 1\), the series converges to \[ \frac{1}{1-r}, \]while if \(|r| \geq 1\), the series diverges.

• Converges if \(|r| < 1\)
• Diverges if \(|r| \geq 1\)

The simplicity of determining convergence for geometric series makes them very useful as comparison tools, particularly in problems utilizing the Direct Comparison Test.

Inequalities in Series

Inequalities play a crucial role when comparing series, especially under the Direct Comparison Test. When we have two series, say \( \sum a_n \) and \( \sum b_n \), with the condition \( 0 \leq a_n \leq b_n \) for all terms, we can make significant conclusions about their behavior.

If the series \( \sum b_n \) converges and \( a_n \) is less than or equal to \( b_n \) for all \( n \), then \( \sum a_n \) also converges. This comparison via inequality allows us to predict the convergence of complex series by relating them to simpler or well-known ones.

• Use inequalities to compare terms of two series.
• If a larger series converges, the smaller one must also converge when terms are non-negative.

Employing inequalities in this way simplifies convergence determinations, making them a critical part of series analysis.

Comparison Series

In the Direct Comparison Test, a comparison series is a secondary series against which we measure the original series. The trick is choosing a series that has a known convergence behavior, such as a geometric or p-series, to serve as this benchmark.

For example, when analyzing the given series \( \sum_{n=1}^{\infty} \frac{1}{n3^n} \), we choose \( \sum_{n=1}^{\infty} \frac{1}{3^n} \) as the comparison series because it is a geometric series with known convergence properties. Since each term \( \frac{1}{n3^n} \) is less than or equal to \( \frac{1}{3^n} \), and the comparison series converges, we conclude the original series converges too.

• Choose comparison series wisely based on known convergence.
• Ensure the comparison gives a clear term-by-term relation through inequalities.

This method is powerful because it turns the problem of an unfamiliar series into one of analyzing a series whose properties we already know.