Problem 31 - Thomas Calculus

Question

Which of the series in Exercises 13 46 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.) $$ \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \) diverges because it is a geometric series with a common ratio greater than 1.

Step 1: Identify the Series

The series given is \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \). Each term of the series is \( \frac{1}{(\ln 2)^{n}} \), where \( \ln 2 \) is a constant (approximately 0.693). This type of series resembles a geometric series.

Step 2: Determine if the Series is Geometric

A geometric series is of the form \( \sum_{n=0}^{\infty} ar^n \) for \( n \geq 0 \), where \( a \) is the first term and \( r \) is the common ratio. For the series \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \), we can express it as a geometric series starting from \( n=1 \) with \( a = \frac{1}{\ln 2} \) and common ratio \( r = \frac{1}{\ln 2} \).

Step 3: Identify Common Ratio and Test for Convergence

For convergence of a geometric series, the absolute value of the common ratio, \( |r| \), must be less than 1. In this series, \( r = \frac{1}{\ln 2} \approx 1.4427 \), which is greater than 1.

Step 4: Conclude the Series Divergence

Since the common ratio \( r \) of the geometric series is greater than 1 (\( \frac{1}{\ln 2} > 1 \)), the series diverges by the geometric series test.

Key Concepts

Geometric Series

When we talk about a geometric series, we refer to a special type of series where each term is obtained by multiplying the previous term by a fixed number, known as the common ratio. This specific pattern makes it quite easy to analyze such series. A geometric series can be expressed in the form: \[ \sum_{n=0}^{\infty} ar^n \] where:

• \(a\) is the first term in the series,
• \(r\) is the common ratio.

For a geometric series to converge, the absolute value of the common ratio \(|r|\) must be less than 1. Convergence means that as you continue to add up or "sum" the terms of the series, they approach a finite limit.
When the common ratio is greater than or equal to 1 in absolute terms, the series does not settle to a finite number, causing the series to diverge. This is a fundamental property to keep in mind when assessing any series with a repeating factor.

Divergence

Divergence is a term used to describe a series that does not converge. In other words, the terms of the series, when summed endlessly, do not approach a finite limit. Instead, the total can grow indefinitely or fluctuate without settling.
In the example of the series \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \), where the common ratio \( r = \frac{1}{\ln 2} \approx 1.4427 \), the series diverges. This is because the common ratio is greater than 1.
This characteristic of a geometric series dictates that when \(|r| \geq 1\), each successive term does not shrink towards zero rapidly enough for the sum to stabilize at a particular value. Diverging series often increase without bounds or fail to approach a single limit.

Convergence Tests

Convergence tests are tools we use to determine if a series converges or diverges. There are several tests available, each suitable to different types of series. For geometric series, the convergence test is straightforward: it revolves primarily around the common ratio.For our specific type of series, the absolute value of the common ratio \(|r|\) has a powerful implication:

• If \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.

These rules derive from the properties of exponential decay and growth respectively.
In more complex situations, other tests like the Ratio Test, Root Test, or Integral Test might be applied. Each of these has varying conditions and outcomes, but they all help in concluding about the nature of the series. Understanding these tests is crucial to effectively analyzing whether infinite series can be summed to a finite amount.