Problem 38 - Thomas Calculus

Question

In Exercises \(31-38,\) use the \(n\) th-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. $$\sum_{n=0}^{\infty} \cos n \pi$$

Step-by-Step Solution

Answer

The series \( \sum_{n=0}^{\infty} \cos(n\pi) \) diverges by the nth-term test.

Step 1: Understand the Test

The nth-term test for divergence states that if \( \lim_{{n \to \infty}} a_n eq 0\), then the series \( \sum_{n=1}^{\infty} a_n \) is divergent. However, if \( \lim_{{n \to \infty}} a_n = 0\), the test is inconclusive.

Step 2: Identify the Sequence

For the given series \( \sum_{n=0}^{\infty} \cos(n\pi) \), the sequence is \( a_n = \cos(n\pi) \). Observe that \( \cos(n\pi) \) alternates between 1 and -1 with every increment of \( n \), due to the periodic nature of the cosine function.

Step 3: Evaluate the Limit

Evaluate the limit: \( \lim_{{n \to \infty}} \cos(n\pi) \). Since \( \cos(n\pi) = (-1)^n \), the limit does not settle to a single value; it alternates between 1 and -1 as \( n \) increases.

Step 4: Apply the nth-Term Test

Since \( \lim_{{n \to \infty}} \cos(n\pi) \) does not equal 0, by the nth-term test for divergence, the series \( \sum_{n=0}^{\infty} \cos(n\pi) \) diverges.

Key Concepts

Sequence

A sequence is a set of numbers arranged in a specific order, following a particular rule. Each number in a sequence is referred to as a term. In the context of the exercise, the sequence is derived from the expression \(a_n = \cos(n\pi)\). Here, each term of the sequence is determined by the cosine of \(n\) multiplied by \(\pi\).

In this specific sequence, there is a special alternation due to the nature of the cosine function:

• When \(n\) is even, \(\cos(n\pi) = 1\).
• When \(n\) is odd, \(\cos(n\pi) = -1\).

The sequence thus alternates between 1 and -1 as \(n\) increases. Understanding this alternation is crucial to analyzing the behavior of the series that it forms.

Infinite Series

An infinite series is formed by adding together all the terms of an infinite sequence. Symbolically, it is represented by the summation notation \(\sum_{n=0}^{\infty} a_n\), where \(a_n\) are the terms in a sequence.

In our exercise, the infinite series is \(\sum_{n=0}^{\infty} \cos(n\pi)\). This series consists of the sum of an infinite number of terms that alternate between 1 and -1.

• Infinite series can either converge to a specific value or diverge.
• The behavior depends on the properties of the sequence terms \(a_n\).

Recognizing patterns in the sequence enables us to better understand the overall tendencies of the series, which is essential for applying tests like the nth-Term Test for Divergence.

Limit

A limit is a fundamental concept in calculus that describes the value a sequence or function approaches as the input approaches some value. It is often represented as \( \lim_{{n \to \infty}} a_n \).

In the context of our exercise, we are tasked to find the limit of the sequence term \(\cos(n\pi)\) as \(n\) approaches infinity. Because \(\cos(n\pi)\) represents \((-1)^n\), it toggles between values 1 and -1:

• If \(n\) approaches infinity, the values continue to oscillate between 1 and -1.
• The limit \( \lim_{{n \to \infty}} \cos(n\pi) \) does not exist in the traditional sense since it does not settle at a single value.

Knowing how to evaluate these limits allows for determining whether or not the series diverges, using specific tests designed to identify such behavior.

Divergent Series

A divergent series is an infinite series that does not converge to a finite limit. This means that the sum of its terms keeps increasing indefinitely or oscillates without approaching a specific value.

In our problem, the series \(\sum_{n=0}^{\infty} \cos(n\pi)\) is a divergent series since the nth-term test for divergence indicates that:

• The limit \( \lim_{{n \to \infty}} \cos(n\pi) eq 0 \).
• The sequence of terms \(\cos(n\pi)\) does not diminish towards zero.

According to the nth-term test, if the limit of the sequence terms does not equal zero, the series is divergent. This means that as you sum over the terms forever, you'll not get closer to a single, finite number, but instead continue to have values that alternate or grow without bound.