Problem 38 - 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.) $$ \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} \) converges by the comparison test.

Step 1: Identify the series

The given series is \( \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} \). It is an infinite series starting from \( n=1 \). We need to determine if it converges or diverges.

Step 2: Analyze the general term

Let's consider the general term of the series, \( a_n = \frac{2}{1+e^{n}} \). As \( n \) increases, \( e^{n} \) becomes very large, making the denominator \( 1+e^{n} \) very large. This implies that \( a_n \to 0 \) as \( n \to \infty \). However, this condition alone does not guarantee convergence.

Step 3: Use the comparison test

To determine convergence or divergence, we use the comparison test. Notice \( \frac{1}{e^n} < \frac{2}{1+e^{n}} < \frac{2}{e^n} \) for sufficiently large \( n \). The series \( \sum_{n=1}^{\infty} \frac{1}{e^{n}} \) is a geometric series with common ratio \( r = \frac{1}{e} < 1 \) and converges. Thus, the comparison test shows that our series also converges.

Key Concepts

Infinite Series

An infinite series is a way to sum an endless sequence of numbers, where the series is expressed in the form \( \sum_{n=1}^{\infty} a_n \). Each number in the sequence is called a term, denoted as \( a_n \). These terms can provide valuable insight into whether or not the series converges to a specific value or diverges. - **Convergence:** If the sum of the terms approaches a specific value as you consider more and more terms, the series is said to converge. This means adding more terms doesn't significantly change the total value. - **Divergence:** On the other hand, if the sum does not approach any finite value, no matter how many terms you add, the series diverges.In the context of the given series \( \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} \), we are exploring whether the series converges by looking at the behavior of its terms as \( n \) approaches infinity.

Comparison Test

The comparison test is a useful tool in determining the convergence or divergence of an infinite series. It involves comparing our series to another one that we already know converges or diverges. - **Basic Principle:** If you have two series \( \sum a_n \) and \( \sum b_n \), and \( 0 \leq a_n \leq b_n \) for all \( n \) beyond some point, then: - If \( \sum b_n \) converges, then \( \sum a_n \) also converges. - If \( \sum a_n \) diverges, then \( \sum b_n \) also diverges.In our exercise, we use the comparison test by noting that \( \frac{1}{e^n} \leq \frac{2}{1+e^{n}} \leq \frac{2}{e^n} \) for large \( n \). Because the series \( \sum_{n=1}^{\infty} \frac{1}{e^n} \) is known to be a convergent geometric series, this helps us conclude that the original infinite series also converges.

Geometric Series

A geometric series is a specific type of infinite series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. A geometric series is given by the formula \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term, and \( r \) is the common ratio.- **Convergence Criteria:** A geometric series converges if the absolute value of the common ratio \( |r| \) is less than 1. If \( |r| \geq 1 \), the series diverges. For example, consider the series \( \sum_{n=1}^{\infty} \frac{1}{e^n} \). This is a geometric series with \( a = \frac{1}{e} \) and the common ratio \( r = \frac{1}{e} \), where \( \frac{1}{e} \) is less than 1. Therefore, this series converges.Understanding this helps in utilizing the comparison test, where comparing with a known convergent geometric series aids in determining the convergence status of another complex series.