Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 is divergent by the divergence test.

Step 1: Rewrite the Series

Rewrite the series in a more manageable form: \[ s_n = n \sin\left(\frac{1}{n}\right). \] We need to analyze the properties of \( s_n \) as \( n \to \infty \).

Step 2: Analyze \( \sin\left(\frac{1}{n}\right) \) as \( n \to \infty \)

The function \( \sin(x) \) is approximately \( x \) when \( x \) is near zero. Therefore, for large \( n \), we have:\[ \sin\left(\frac{1}{n}\right) \approx \frac{1}{n}. \] This approximation implies:\[ n \sin\left(\frac{1}{n}\right) \approx n \cdot \frac{1}{n} = 1. \] Thus, the general term does not approach zero as \( n \to \infty \).

Step 3: Apply the Divergence Test

Since the limit of the sequence \( s_n = n \sin\left(\frac{1}{n}\right) \) as \( n \to \infty \) is 1 and not zero:\[ \lim_{{n \to \infty}} s_n = 1 eq 0, \]we can conclude, by the divergence test, that the series:\[ \sum_{n=1}^{\infty} n \sin\left(\frac{1}{n}\right) \] is divergent.

Key Concepts

Divergence Test

The divergence test, also known as the nth-term test for divergence, is one of the simplest tests for series convergence. Its purpose is to check whether the terms of a series do not approach zero as the term number increases indefinitely. Here is how it works in a nutshell:

For a series \( \sum a_n \), if \( \lim_{{n \to \infty}} a_n eq 0 \), then the series is divergent. This means the terms do not become arbitrarily small, signaling that the series does not converge. However, if the limit is 0, this does not guarantee convergence; other tests are needed.

In the case of the series \( \sum_{n=1}^{\infty} n \sin\left(\frac{1}{n}\right) \), we found \( \lim_{{n \to \infty}} n \sin\left(\frac{1}{n}\right) = 1 \). Since this limit isn't zero, the divergence test confirms the series diverges.

Sin Function Behavior

Understanding the behavior of the sine function becomes crucial, especially when analyzing limits. The sine function, \( \sin(x) \), behaves predictably near zero, which is where the approximation becomes useful. Let's explore this concept.

• Close to zero, \( \sin(x) \approx x \). This approximation holds because the sine function is linear around zero, with \( \sin(x) \) closely matching the line \( y = x \).
• When substituting \( x = \frac{1}{n} \), and considering large \( n \), \( \sin\left(\frac{1}{n}\right) \approx \frac{1}{n} \). Therefore, multiplying \( n \) by \( \sin\left(\frac{1}{n}\right) \) results in \( 1 \), explaining why the earlier terms do not shrink to zero.

This fundamental property helps immensely in series evaluation, as it simplifies our understanding, particularly in conjunction with limit processes.

Limit Comparison

While the solution used the divergence test decisively, it's useful to know about the limit comparison test as an alternative when exploring series convergence. The limit comparison test assesses the behavior of two series together. Here’s a quick overview:

• Suppose \( \{a_n\} \) and \( \{b_n\} \) are series with positive terms. If \( \lim_{{n \to \infty}} \frac{a_n}{b_n} = c \) where \( c \) is a positive finite number, both series either converge or diverge together.
• This idea is particularly helpful when direct analysis of the nth-term is complicated, allowing us to relate it to a simpler series.

Although not necessary for this exercise, understanding various techniques enriches comprehension and opens more paths for analyzing convergence efficiently.