Problem 40 - Calculus: Early Transcendentals

Question

A series \( \sum a_n \) is defined by the equations \( a_1 = 1 \) \( a_{n+1} = \frac {2 + \cos n}{\sqrt{n}} a_n \) Determine whether \( \sum a_n \) converges or diverges.

Step-by-Step Solution

Answer

The series \( \sum a_n \) diverges.

Step 1: Understand the Recursive Formula

The series is defined by a recursive formula: \( a_1 = 1 \) and \( a_{n+1} = \frac{2 + \cos n}{\sqrt{n}} a_n \). We need to determine if this series \( \sum a_n \) converges or diverges.

Step 2: Examine the Growth of Terms

To gain insight into the behavior of \( a_n \), consider the factor \( \frac{2 + \cos n}{\sqrt{n}} \). As \( n \) increases, \( \cos n \) oscillates between -1 and 1. Hence, \( 1 \leq 2 + \cos n \leq 3 \). Thus, \( \frac{2 + \cos n}{\sqrt{n}} \) behaves as \( \frac{C}{\sqrt{n}} \) for some constant \( C \) between 1 and 3.

Step 3: Apply the Test for Convergence

Given \( \frac{1}{\sqrt{n}} \) decreases to zero as \( n \to \infty \), we compare our term \( a_n = a_{n-1} \times \frac{2 + \cos(n-1)}{\sqrt{n-1}} \) to the harmonic series argument for divergence. Since \( a_1 = 1 \), subsequent \( a_n \) decrease but not fast enough to zero. The conditions for conditional convergence are not met.

Step 4: Conclude the Behavior of the Series

The general term \( a_n \sim \frac{1}{\sqrt{n!}} \) implies \( \sum a_n \) resembles \( \sum \frac{1}{n^{3/2}} \), which is a divergent p-series when \( p \leq 1 \). Thus, the series diverges.

Key Concepts

Understanding Divergence

In the context of series, when we talk about divergence, we are referring to series whose terms grow to such an extent that their sum doesn't settle or approach a finite number. When we analyzed the series from the problem, we found that as we consider more terms, the sequence doesn't "calm down" or sum up to a fixed limit. Instead, it keeps growing.
This growth suggests that the series lacks the properties needed for convergence, i.e., the sums don't add up to a specific point but keep expanding. Key indicators of divergence include terms that, although getting smaller, do not trend to zero speedily enough; or series that, even with zeroing terms, accumulate to infinity. That's precisely the situation we encountered with our series.

The Role of the Recursive Formula

A recursive formula allows each term in a series to be defined based on previous terms. In our situation, we have the recursive rule:

• \( a_1 = 1 \)
• \( a_{n+1} = \frac{2 + \cos n}{\sqrt{n}} a_n \)

In simpler terms, once you know a starting point (here, \( a_1 = 1 \)), each subsequent term can be calculated using the formula. This recursive relationship is central in understanding the growth of terms.
Each term builds on the last, and how it scales is crucial. The term \( \frac{2 + \cos n}{\sqrt{n}} \) dictates how \( a_n \) behaves as \( n \) advances. Recognizing how the switch in \( n \) influences each term's size helps decode the series' behavior. Recursive formulas contrast with explicit formulas, where terms stand independently, and give more insight into the sequence dynamics.

The Nature of Harmonic Series

Harmonic series are classical examples in mathematics where every term is inverse of its corresponding positive integer. This series is defined as:

• \( \sum_{n=1}^{\infty} \frac{1}{n} \)

In our exercise exploration, we compared the derived terms \( a_n \) to the harmonic series structure to gain a sense of their divergence. Even though terms decline towards zero, their collective sum tends challenges intuition as they don't reach a finite limit.
The significance of the harmonic series appears largely from its role as a divergence benchmark: it's divergent despite having terms that decrease to zero, teaching that mere shrinkage isn't sufficient for convergence. Every student exploring properties of series must familiarise themselves with this foundational paradigm.

Identifying with P-Series

To further analyze series like our given problem, we often leverage the concept of a p-series. A p-series takes form as:

• \( \sum_{n=1}^{\infty} \frac{1}{n^p} \)

where \( p \) is a positive constant. For convergence assessment, the value of \( p \) is vital; specifically, the series converges if \( p > 1 \) and diverges otherwise.
In our step-by-step solution, the series comparison depended on this critical insight. Observing that our terms behaved similarly to a p-series with \( p \leq 1 \) supported the conclusion of divergence. For learners, understanding p-series guides not only in knowing when series converge, but also when they can't, setting pivotal benchmarks for series analysis.