Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} n \sin \left(\frac{1}{n}\right) $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Understanding the Series

The given series is \( \sum_{n=1}^{\infty} n \sin \left(\frac{1}{n}\right) \). To classify it, we need to determine if it converges absolutely, conditionally, or if it diverges.

Step 2: Analyzing the Terms

We consider the behavior of the sequence \( a_n = n \sin \left(\frac{1}{n}\right) \) as \( n \to \infty \). A useful approximation when \( x \) is small is \( \sin x \approx x \). Thus, \( \sin \left(\frac{1}{n}\right) \approx n \cdot \frac{1}{n} = 1 \).

Step 3: Applying the Divergence Test

The Divergence Test says that if \( \lim_{n \to \infty} a_n eq 0 \), then \( \sum a_n \) diverges. Since \( a_n \approx 1 \) for large \( n \), we have \( \lim_{n \to \infty} a_n = 1 \).

Step 4: Concluding Divergence

Since \( \lim_{n \to \infty} a_n = 1 eq 0 \), by the Divergence Test, the series \( \sum_{n=1}^{\infty} n \sin \left(\frac{1}{n}\right) \) diverges.

Key Concepts

Understanding the Divergence Test

The divergence test is a fundamental tool in analyzing the convergence of series. It involves examining the limit of the sequence that makes up the series. Here’s how it works:

• If the limit of the terms as n approaches infinity is not zero, then the series must diverge.
• If the limit is zero, the test is inconclusive, and further investigation is needed to determine convergence or divergence.

For the series \( \sum_{n=1}^{\infty} n \sin \left(\frac{1}{n}\right) \), the terms become approximately 1 as \(n\) becomes very large. Since the limit of these terms is 1, not 0, the divergence test quickly informs us that this series diverges.

Exploring Trigonometric Series

Trigonometric series, formed from trigonometric functions such as sine and cosine, can be complex to evaluate. However, they often have special properties and patterns that can be used to evaluate convergence.In our series, each term is \( n \sin \left( \frac{1}{n} \right) \). The sine function is part of the series, making it a trigonometric series. For small values of \( x \), it is useful to remember that \( \sin x \approx x \), which simplifies analysis.In this series, \( \sin \left(\frac{1}{n}\right) \approx \frac{1}{n} \). Since each term simplifies to approximately \( n \cdot \frac{1}{n} = 1 \), we apply this approximation to gain insights into its convergence, although further steps like the divergence test are necessary to make conclusions.

Understanding Limit Comparison

The limit comparison test is another powerful method used to determine the convergence or divergence of series. It involves comparing a complicated series to a simpler, often well-known series. To use this test:

• Select \( b_n \) as a series you know the behavior of.
• Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If this limit is a positive finite number, then both series \( \sum a_n \) and \( \sum b_n \) behave similarly.

For a series like \( \sum_{n = 1}^{\infty} n \sin \left( \frac{1}{n} \right) \), limit comparison can be employed, although here the divergence test suffices.

Understanding Infinite Series Convergence

Infinite series convergence is a cornerstone of calculus, vital for understanding various mathematical concepts. Convergence means the terms of the series approach a particular value, providing a finite sum. There are a few ways in which series can differ:

• Absolutely Convergent: If \( \sum |a_n| \) converges, then so does \( \sum a_n \).
• Conditionally Convergent: \( \sum a_n \) converges but \( \sum |a_n| \) does not.
• Divergent: Neither \( \sum a_n \) nor \( \sum |a_n| \) converges.

Understanding the classification helps distinguish series and analyze them effectively. For our series, finding it neither converges absolutely nor conditionally, but diverges, helped in completing the task.