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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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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06:52

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.

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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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