Ch. 10 - Infinite Sequences and Series

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 10 - Infinite Sequences and Series

Problem 10.4.66

Chapter 10, Problem 10.4.66

If ∑aₙ is a convergent series of positive terms, prove that ∑sin(aₙ) converges.

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Recall that since \( \sum a_n \) is a convergent series with positive terms, we have \( a_n \to 0 \) as \( n \to \infty \). This is a key fact to use when analyzing \( \sin(a_n) \).

Use the fact that for small values of \( x \), \( \sin x \) behaves like \( x \). More precisely, by the Taylor expansion, \( \sin x = x - \frac{x^3}{6} + O(x^5) \). This implies that \( \sin(a_n) \sim a_n \) when \( a_n \) is close to zero.

Since \( a_n > 0 \) and \( a_n \to 0 \), for sufficiently large \( n \), \( \sin(a_n) \) is positive and approximately equal to \( a_n \). Therefore, there exists some \( N \) such that for all \( n > N \), \( 0 < \sin(a_n) \leq a_n \).

Apply the Comparison Test for series: because \( 0 < \sin(a_n) \leq a_n \) for large \( n \), and \( \sum a_n \) converges, it follows that \( \sum \sin(a_n) \) also converges.

Conclude that the series \( \sum \sin(a_n) \) converges by the Comparison Test, using the fact that \( \sin(a_n) \) behaves like \( a_n \) for small positive terms and \( \sum a_n \) converges.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Convergence of Series with Positive Terms

A series ∑aₙ with positive terms converges if the sequence of partial sums approaches a finite limit. This implies that the terms aₙ themselves approach zero as n approaches infinity, which is crucial for analyzing related series.

Behavior of the Sine Function for Small Arguments

For small values of x, sin(x) is approximately equal to x, since sin(x) ~ x - x³/6 + ... . This approximation allows us to compare sin(aₙ) to aₙ when aₙ is small, facilitating the use of comparison tests for convergence.

Comparison Test for Series Convergence

The comparison test states that if 0 ≤ bₙ ≤ cₙ for all n beyond some index and ∑cₙ converges, then ∑bₙ also converges. Applying this test by comparing sin(aₙ) to aₙ helps prove the convergence of ∑sin(aₙ) given the convergence of ∑aₙ.

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Textbook Question

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