Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13-22, 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 Divergence Test, as the terms do not tend to zero.

Step 1: Identify the given series

The series we are asked to consider is \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \). Our task is to determine whether this series converges or diverges.

Step 2: Explore the behavior of the series terms

Consider the behavior of the individual terms. We know that as \( k \to \infty \), \( \frac{1}{k} \to 0 \), thus \( \sin \frac{1}{k} \to \sin 0 = 0 \). However, \( \sin x \approx x \) when \( x \) is near zero, so \( \sin \frac{1}{k} \approx \frac{1}{k} \). Therefore, \( k \sin \frac{1}{k} \approx k \cdot \frac{1}{k} = 1 \). This suggests the terms do not tend to zero as \( k \to \infty \).

Step 3: Apply the Divergence Test

The Divergence Test (also known as the nth-term test for divergence) states that if \( \lim_{k \to \infty} a_k eq 0 \), then \( \sum_{k=1}^{\infty} a_k \) diverges. Here, \( a_k = k \sin \frac{1}{k} \approx 1 \). Therefore, \( \lim_{k \to \infty} a_k = 1 eq 0 \).

Step 4: Conclusion from the Divergence Test

Since the limit of the terms is not zero, by the Divergence Test, the series \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \) diverges.

Key Concepts

Divergence Test

The Divergence Test is an essential tool in calculus for determining the behavior of series. It's also known as the nth-term test for divergence. The test is powerful yet simple: if the limit of the series terms does not approach zero, the series must diverge.
Consider a series with terms \( a_k \). To apply the Divergence Test, calculate the limit \( \lim_{k \to \infty} a_k \).

• If \( \lim_{k \to \infty} a_k eq 0 \), then the series diverges.
• If \( \lim_{k \to \infty} a_k = 0 \), the test is inconclusive; the series could converge or diverge, and further tests are needed.

It's essential to apply this test first as it's a quick way to identify divergence without complex calculations. For the series \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \), since the terms approach 1 (not zero), the series diverges. This straightforward yet crucial test can save time and ensure that a series doesn't mistakenly seem convergent when it's not.

Sin Approximation

The sin function plays a vital role in many mathematical calculations, and understanding how to approximate it can simplify complex problems. Around zero, the sin function behaves almost linearly.
When dealing with small angles \( x \),\[ \sin x \approx x \]This approximation is especially useful in calculus when examining the behavior of functions as they approach zero.
In the series \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \), as \( k \) grows large, \( \frac{1}{k} \) becomes very small, nearing zero.
Thus, we use the approximation of \( \sin \frac{1}{k} \approx \frac{1}{k} \) for simplification.

• For very small \( x \), the error is negligible, and this simplification provides a clear view of the series behavior.

Understanding these approximations is crucial for effective calculus problem-solving since they can reduce the complexity of solving or simplifying series.

Series Divergence

Series divergence is a central topic in calculus, where the goal is to determine whether an infinite series converges to a finite value or not. Divergence means the series does not settle at a fixed value.
When analyzing a series, look for signs of divergence, such as non-zero limits or growing terms.

• In the series \( \sum_{k=1}^{\infty} k \sin \frac{1}{k} \), the approximation \( k \sin \frac{1}{k} \approx 1 \) indicates terms remain constant as \( k \) increases, making it unlikely to converge.
• Such series, where terms fail to approach zero, or indicate growth, signal divergence.

Recognizing divergence early can guide you to use other tests for confirmation or search for convergence in more complex scenarios. Grasping these concepts can help tackle a variety of calculus problems effectively.

Calculus Problem-Solving

Calculus problem-solving requires a strategic approach to analyze and break down complex mathematical problems. Identifying appropriate tests and using mathematical approximations are key.
Start by understanding the problem components and experiment with different techniques:

• Explore limits and behavior of terms, using approximations like \( \sin x \approx x \) for simplification.
• Apply specific convergence or divergence tests, such as the Divergence Test, which focuses on the behavior of individual terms.

Also, use these problem-solving strategies:

• Visualize the problem by perhaps graphing the function or considering numerical limits.
• Cross-reference results with known series to intuit their behavior.

By combining these tested techniques, addressing detailed calculus problems becomes less daunting, aiding in accurate and efficient solutions.