Problem 46 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = 2^{-n} \cos n \pi \)

Step-by-Step Solution

Answer

The sequence converges; its limit is 0.

Step 1: Determine the Behavior of the Terms

The sequence is given by \( a_n = 2^{-n} \cos(n\pi) \). We need to determine the behavior of the sequence by examining \( 2^{-n} \) and \( \cos(n\pi) \) separately.

Step 2: Analyze \( 2^{-n} \)

The term \( 2^{-n} \) can be rewritten as \( \frac{1}{2^n} \). As \( n \to \infty \), \( \frac{1}{2^n} \to 0 \) since \( 2^n \) grows exponentially and dominates the numerator, leading to zero.

Step 3: Analyze \( \cos(n\pi) \)

The term \( \cos(n\pi) \) will alternate between 1 and -1 as \( n \) takes integer values. Specifically, if \( n \) is even, \( \cos(n\pi) = 1 \), and if \( n \) is odd, \( \cos(n\pi) = -1 \). Therefore, \( \cos(n\pi) \) does not converge but oscillates between two values.

Step 4: Combine the Results

The sequence is \( a_n = 2^{-n} \cos(n\pi) \). Combine the behavior of both components. Since \( 2^{-n} \to 0 \) and \( \cos(n\pi) \) oscillates between -1 and 1, the product \( 2^{-n} \cos(n\pi) \) results in terms that get increasingly small in magnitude, leading \( \lvert a_n \rvert \leq 2^{-n} \). As \( n \to \infty \), this product approaches zero, demonstrating convergence.

Step 5: Conclusion

Given that the magnitude of \( 2^{-n} \cos(n\pi) \) approaches zero, the sequence \( a_n = 2^{-n} \cos(n\pi) \) converges. The limit of the sequence is 0.

Key Concepts

Understanding the Limit of a Sequence

The limit of a sequence is a core concept in calculus and mathematical analysis. It describes the behavior of a sequence as the index approaches infinity. To determine if a sequence has a limit, we check if its terms approach a specific number as the index increases indefinitely.

For example, consider the sequence described by formula \( a_n = 2^{-n} \). As \( n \to \infty \), each term \(2^{-n}\) becomes smaller and closer to zero. This means that the limit of the sequence \( 2^{-n} \) is 0.

• This pattern shows that the sequence approaches 0 as \( n \) increases, confirming that it converges to a limit.
• If a sequence has a limit, it is called convergent; otherwise, it is termed divergent.

Understanding limits helps us analyze the long-term behavior of sequences and functions, providing insights into their stability and potential applications.

The Concept of a Convergent Series

A convergent series is related to the summation of terms in a sequence. When the sum of an infinite series approaches a specific value, we say the series converges.

An example can be seen with the geometric series \( \sum_{n=0}^{\infty} ar^n \), where \( |r| < 1 \). Such a series will sum to \( \frac{a}{1-r} \) as \( n \to \infty \). This sum is finite and shows that the series converges.

• In our exercise, the sequence itself is \( a_n = 2^{-n} \cos(n\pi) \).
• Its individual terms converge to 0, indicating that its series, if summed, would also have properties indicating convergence.

The understanding of convergent series is crucial as it allows mathematicians to infer properties about infinite processes, leading to many significant conclusions in various fields of study.

Discovering Oscillating Sequences

An oscillating sequence is one whose terms do not settle down to a specific number but instead fluctuate between two or more values. A typical example is the sequence \( (-1)^n \), which oscillates between -1 and 1.

In the given problem, the term \( \cos(n\pi) \) within the sequence \( a_n = 2^{-n} \cos(n\pi) \) exhibits oscillation. As \( n \) alternates between even and odd, \( \cos(n\pi) \) changes from 1 to -1 and back.

• This oscillation means \( \cos(n\pi) \) itself does not converge, although when it's combined with \( 2^{-n} \), the resulting product converges to 0.
• Recognizing oscillation is essential in determining the broader behavior of sequences.

While oscillating sequences on their own don't converge, when paired wisely with decreasing terms, like in this exercise, they can contribute to a convergent sequence overall.