Problem 30 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{(\ln 3)^{n}} $$

Step-by-Step Solution

Answer

The series converges as it is a geometric series with \( |r| < 1 \).

Step 1: Recognize the series type

The series given is \( \sum_{n=1}^{\infty} \frac{1}{(\ln 3)^{n}} \). This is an example of a geometric series where the general term can be written in the form \( ar^{n-1} \).

Step 2: Identify the parameters for a geometric series

For a geometric series \( \sum_{n=1}^{\infty} ar^{n-1} \), the series converges if the absolute value of the common ratio \( |r| < 1 \). Otherwise, it diverges. Here, \( a = \frac{1}{\ln 3} \) and \( r = \frac{1}{\ln 3} \).

Step 3: Check the common ratio condition

Calculate the value of the common ratio \( r = \frac{1}{\ln 3} \). Since \( \ln 3 \) is approximately 1.0986, we have \( r = \frac{1}{1.0986} \approx 0.9102 \). Therefore, \( |r| < 1 \).

Step 4: Determine convergence or divergence

Since the value \( |r| < 1 \), the series \( \sum_{n=1}^{\infty} \frac{1}{(\ln 3)^{n}} \) is a convergent geometric series.

Key Concepts

Geometric Series

A geometric series is a special type of infinite series where each term in the series is a fixed multiple of the preceding term. This fixed multiple is known as the common ratio. The general form of a geometric series can be written as \( \sum_{n=1}^{\infty} ar^{n-1} \), where \( a \) is the first term, and \( r \) is the common ratio.

To determine whether a given geometric series converges or diverges, one must examine the common ratio \( r \). Specifically, the series converges if the absolute value of \( r \) is less than one (\( |r| < 1 \)).

• Convergent series: If \( |r| < 1 \), the geometric series will converge, meaning it will approach a finite sum.
• Divergent series: If \( |r| \geq 1 \), the series will continue to increase indefinitely and thus diverge.

In the problem solution, the series was identified as a geometric series, which is the first crucial step in assessing whether it converges.

Infinite Series

The concept of an infinite series is fundamental in calculus, representing the sum of an infinite sequence of terms. More formally, an infinite series is expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the terms of the series.

When dealing with infinite series, a key question is whether the series converges, meaning the sum approaches a specific finite value as more terms are added, or diverges as it keeps growing indefinitely.

• Convergence: If an infinite series approaches a finite limit as more terms are added, it is described as convergent.
• Divergence: Conversely, if the sum becomes infinitely large, the series is divergent.

Understanding infinite series is critical to solving many calculus problems, especially when confirming if a series like the one in our original exercise converges using specific tests like the geometric series test.

Common Ratio

The common ratio is a pivotal component in understanding geometric series. It specifies the factor by which each term in the series is multiplied to obtain the next term. In a geometric series \( \sum_{n=1}^{\infty} ar^{n-1} \), \( r \) is this common ratio, critical for evaluating series behavior.

To determine the convergence of a geometric series, the value of the common ratio is examined. If the absolute value of \( r \) is less than one, \( |r| < 1 \), the series converges.

• A common ratio of \( |r| < 1 \) indicates convergence, implying the terms shrink and sum to a finite number.
• If \( |r| \geq 1 \), the terms don’t shrink sufficiently, causing divergence and the sum to grow indefinitely.

In the solution provided, it was found that \( r = \frac{1}{\ln 3} \), approximately 0.9102, leading to a conclusion that the series converges, emphasizing the importance of calculating the common ratio correctly.