Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use any test developed so far, including any from Section \(9.2\), to decide about the convergence or divergence of the series. Give a reason for your conclusion. $$ \sum_{k=1}^{\infty} k \sin \frac{1}{k} $$

Step-by-Step Solution

Answer

The series diverges by the Comparison Test.

Step 1: Analyze the General Term

Let's first examine the general term of the series: \( a_k = k \sin \left( \frac{1}{k} \right) \). We notice that as \( k \to \infty \), \( \frac{1}{k} \to 0 \). This prompts us to consider the behavior of \( \sin \left( \frac{1}{k} \right) \). Utilizing the approximation \( \sin x \approx x \) when \( x \to 0 \), we approximate: \( \sin \left( \frac{1}{k} \right) \approx \frac{1}{k} \). Thus, the term \( a_k \approx 1 \).

Step 2: Test for Convergence Using Comparison Test

Since \( a_k \approx 1 \), the series resembles \( \sum_{k=1}^{\infty} 1 \), which is a divergent harmonic series (it diverges because the terms do not tend to zero as \( k \to \infty \)). Since the general term \( a_k \approx 1 \) also does not approach zero, we suspect divergence. Let's use the Comparison Test: if \( b_k = 1 \) for all \( k \), then \( 0 < a_k < b_k \), and since \( \sum b_k \) diverges, \( \sum a_k \) must also diverge.

Step 3: Conclude Based on the Test

The series \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \) diverges because the terms of the series \( a_k \) do not tend to zero and are comparable to a known divergent series. This conclusion is supported by both the limit approximation and the Comparison Test.

Key Concepts

Comparison Test

The Comparison Test is a powerful tool for determining the convergence or divergence of infinite series. It works by comparing a given series to a second series with known convergence or divergence behavior. Here's the basic idea:

• If a series \( \sum a_k \) and another series \( \sum b_k \) satisfy \( 0 \leq a_k \leq b_k \) for all \( k \),
• and since \( \sum b_k \) diverges, then \( \sum a_k \) must also diverge.
• Conversely, if \( \sum b_k \) converges, then \( \sum a_k \) would converge as well.

In our exercise, we use the comparison test by approximating \( a_k \approx 1 \), directly comparing it to the harmonic series \( \sum_{k=1}^{\infty} 1 \). Since the harmonic series diverges, so must our series based on this approximation.

Approximation of Sine Function

When evaluating the behavior near zero, the sine function can be approximated using a simple formula. For small angles, we have:

• \( \sin x \approx x \) when \( x \to 0 \).

This approximation is crucial in analyzing series where sine functions appear with small arguments, like in our problem:

• For \( \sin \left( \frac{1}{k} \right) \), since \( \frac{1}{k} \to 0 \) and is small, the approximation \( \sin \left( \frac{1}{k} \right) \approx \frac{1}{k} \) holds.

Substituting this into the general term \( k \sin \left( \frac{1}{k} \right) \) gives \( a_k \approx 1 \). This leads us to model the series similarly to well-known divergent series, aiding in our analysis of convergence or divergence.

Harmonic Series Divergence

The harmonic series is a classic example of a divergent series. It is expressed as:

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

Despite its terms tending to zero, the series itself diverges. This is one of the key learning points in understanding series behavior:

• Just because terms tend to zero, does not guarantee series convergence.

The divergence of the harmonic series is essential in many mathematical tests and approximations. In our exercise, the behavior of \( a_k \approx 1 \) brings us to compare with the harmonic series, leading to the conclusion that the original series diverges. Understanding the conditions and implications of the harmonic series divergence aids in making insightful conclusions about other series.