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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.4.45
Chapter 10, Problem 10.4.45

Determining Convergence or Divergence
Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.
∑ (from n=1 to ∞) sin (1/n)

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Recognize that the series is \( \sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right) \). Our goal is to determine whether this infinite series converges or diverges.
Recall the behavior of \( \sin x \) near zero: for small \( x \), \( \sin x \approx x \). Since \( \frac{1}{n} \to 0 \) as \( n \to \infty \), we can approximate \( \sin\left(\frac{1}{n}\right) \approx \frac{1}{n} \) for large \( n \).
Compare the given series to the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is a well-known divergent series. Since \( \sin\left(\frac{1}{n}\right) \) behaves like \( \frac{1}{n} \) for large \( n \), the terms do not decrease fast enough to guarantee convergence.
Use the Limit Comparison Test by evaluating \( \lim_{n \to \infty} \frac{\sin(1/n)}{1/n} \). If this limit is a finite nonzero number, then both series either converge or diverge together.
Since the harmonic series diverges and the limit comparison test shows similar behavior, conclude that the series \( \sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right) \) diverges.

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Key Concepts

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

Convergence and Divergence of Infinite Series

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely; otherwise, it diverges. Understanding this concept is fundamental to analyzing whether a given series sums to a finite value or not.
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Convergence of an Infinite Series

Limit Comparison and Behavior of Terms

Examining the behavior of the terms as n approaches infinity helps determine convergence. If the terms do not approach zero, the series diverges. For example, since sin(1/n) ~ 1/n for large n, comparing with the harmonic series is useful.
Recommended video:
07:45
Limit Comparison Test

Comparison Test and Asymptotic Approximations

The Comparison Test involves comparing a given series to a known benchmark series to infer convergence or divergence. Using asymptotic approximations like sin(1/n) ≈ 1/n for large n allows applying this test effectively.
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Direct Comparison Test
Related Practice
Textbook Question

In Exercises 125–134, determine whether the sequence is monotonic, whether it is bounded, and whether it converges.

aₙ = (4ⁿ⁺¹ + 3ⁿ) / 4ⁿ

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Textbook Question

Direct Comparison Test

In Exercises 1–8, use the Direct Comparison Test to determine if each series converges or diverges.

∑ (from n=1 to ∞) (√n + 1) / (√(n² + 3))

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Textbook Question

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑ (from n = 2 to ∞) [(ln n / n)³]

Textbook Question

Recursively Defined Sequences

In Exercises 101–108, assume that each sequence converges and find its limit.

a₁ = 5,aₙ₊₁ = √(5aₙ)

Textbook Question

In Exercises 37–42, find the series’ radius of convergence.

∑ (from n = 1 to ∞) [ (n!)² / (2ⁿ (2n)!) ] xⁿ

Textbook Question

Finding nth Partial Sums

In Exercises 1–6, find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.

2 + (2/3) + (2/9) + (2/27) + … + (2 / 3ⁿ⁻¹) + …