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)

Verified step by step guidance

1. 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.
2. 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 \( n \).
3. 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.
4. 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.
5. 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.

Key Concepts

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.

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.

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.