Problem 66 - Thomas Calculus

Question

If \(\sum a_{n}\) is a convergent series of positive terms, prove that \(\sum \sin \left(a_{n}\right)\) converges.

Step-by-Step Solution

Answer

Since \(\sum a_n\) converges and \(\sin(a_n) \leq a_n\), \(\sum \sin(a_n)\) converges by the Comparison Test.

Step 1: Understanding Convergence

A series \(\sum a_n\) of positive terms converges if its sequences of partial sums \(s_n = a_1 + a_2 + \cdots + a_n\) have a limit as \(n\) approaches infinity. This implies the terms \(a_n\) approach zero.

Step 2: Behavior of Sine for Small Values

Since \(a_n\) approaches zero, for sufficiently small \(a_n\), we can use the approximation \(\sin(a_n) \approx a_n\). Specifically, we can use \(\sin(a_n) \leq a_n\) because \(\sin(x)\) is increasing for small positive \(x\) and \(x\) itself is always an upper bound of \(\sin(x)\).

Step 3: Comparison Test

For convergence, we can utilize the Comparison Test. We already have \(\sin(a_n) \leq a_n\). Since \(\sum a_n\) converges by assumption and the terms \(\sin(a_n)\) are non-negative, by the Comparison Test, the series \(\sum \sin(a_n)\) also converges.

Key Concepts

Comparison Test

The Comparison Test is a valuable tool for determining if a series converges or diverges. It's quite straightforward:

• To use it, you need two series, say \( \sum a_n \) and \( \sum b_n \), both composed of non-negative terms.
• If \( b_n \geq a_n \) for all \( n \) and \( \sum b_n \) is known to converge, then \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \) for all \( n \) and \( \sum a_n \) diverges, then \( \sum b_n \) also diverges.

This test is particularly useful when dealing with complex terms, as it allows us to replace these terms with simpler ones for which we already know the convergence behavior. In the exercise, we compare \( \sum \sin(a_n) \) and \( \sum a_n \) using this principle to establish the convergence of \( \sum \sin(a_n) \).
By knowing that \( \sum a_n \) converges and \( \sin(a_n) \leq a_n \), it's clear that \( \sum \sin(a_n) \) must also converge by the Comparison Test.

Sine Function

The sine function, denoted \( \sin(x) \), is a fundamental trigonometric function. It maps any real number \( x \) to the range \([-1, 1]\). For small positive values of \( x \), such as the terms of a series that converge to zero, \( \sin(x) \) has some interesting properties:

• \( \sin(x) \) is an increasing function, meaning it gets bigger as \( x \) gets bigger.
• For \( x \geq 0 \), we have \( 0 \leq \sin(x) \leq x \).
• This last point is crucial because it allows us to compare \( \sin(a_n) \) directly with \( a_n \) in the exercise.

The sine function's behavior at small values helps in approximating and comparing it with simpler linear functions, which is often useful for assessing convergence.

Approximation of Sine

Approximating \( \sin(x) \) can be incredibly helpful, especially when \( x \) is very small. This approximation is based on the Taylor series expansion around 0:\[ \sin(x) \approx x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \]For small \( x \), the higher-order terms become negligibly small, and it's often sufficient to approximate \( \sin(x) \) by \( x \). This gives us:

• \( \sin(x) \approx x \) for \( x \approx 0 \)
• Since all further terms in the series are positive for very small \( x \), \( \sin(x) \leq x \) always holds for positive \( a_n \).

This approximation leads directly into applying the Comparison Test, as highlighted in the exercise. It provides an upper bound for \( \sin(a_n) \) by \( a_n \), making it possible to use known convergent properties of \( \sum a_n \) to infer those of \( \sum \sin(a_n) \).

Positive Series

A positive series is simply a series where all the terms are non-negative. That is, \( a_n \geq 0 \) for every term in the sequence. This property makes analyzing convergence more straightforward in several ways:

• The partial sums of positive series are non-decreasing, as each additional positive term adds to the previous sum.
• Because positive terms accumulate, the presence or absence of a limit can indicate convergence or divergence.

For convergence, a positive series must approach a finite limit as more terms are added. In our exercise, \( \sum a_n \) is a convergent series of positive terms, leading to the key approximation and application of the Comparison Test to determine the behavior of the series \( \sum \sin(a_n) \). Understanding the nature of positive series helps in using these strategies effectively to solve convergence-related problems.