9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.

∑ (k = 1 to ∞) sin(1 / k) / k²

검증된 단계별 안내

First, identify the general term of the series: \(a_k = \frac{\sin(1/k)}{k^2}\).

Recall that for large \(k\), \(\sin(1/k)\) behaves approximately like \(1/k\) because \(\sin x \approx x\) when \(x\) is close to 0.

Use this approximation to compare \(a_k\) with a simpler series term: \(b_k = \frac{1/k}{k^2} = \frac{1}{k^3}\).

Since \(\sum_{k=1}^\infty \frac{1}{k^3}\) is a p-series with \(p=3 > 1\), it converges.

Apply the Limit Comparison Test by computing \(\lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\sin(1/k)/k^2}{1/k^3} = \lim_{k \to \infty} \frac{\sin(1/k)}{1/k}\). Since this limit is finite and nonzero, both series share the same convergence behavior, so the original series converges.

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주요 개념

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Comparison Test

The Comparison Test determines the convergence of a series by comparing it to a second series with known behavior. If the terms of the given series are smaller than those of a convergent series with positive terms, then the given series also converges. Conversely, if the terms are larger than those of a divergent series, the given series diverges.

Limit Comparison Test

The Limit Comparison Test compares two series by taking the limit of the ratio of their terms. If this limit is a positive finite number, both series either converge or diverge together. This test is useful when direct comparison is difficult but the terms behave similarly for large indices.

Behavior of sin(1/k) for large k

For large values of k, sin(1/k) can be approximated by its argument 1/k because sin(x) ≈ x when x is close to zero. This approximation helps simplify the given series terms to roughly 1/(k³), aiding in the application of comparison tests with p-series.

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.∑ (k = 1 to ∞) sin(1 / k) / k²