42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 3 to ∞)ln(k) / k³ᐟ²

검증된 단계별 안내

Identify the series given: \( \sum_{k=3}^{\infty} \frac{\ln(k)}{k^{3/2}} \). We want to determine if this series converges or diverges.

Consider the behavior of the terms \( a_k = \frac{\ln(k)}{k^{3/2}} \) as \( k \to \infty \). Since \( \ln(k) \) grows slowly and \( k^{3/2} \) grows faster, the terms approach zero, which is necessary for convergence but not sufficient.

Choose a convergence test suitable for series with positive terms and involving logarithms and powers. The Comparison Test or the Limit Comparison Test are good candidates here.

Compare \( a_k \) with a simpler series \( b_k = \frac{1}{k^{3/2}} \), which is a p-series with \( p = \frac{3}{2} > 1 \) and is known to converge.

Apply the Limit Comparison Test by evaluating \( \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\ln(k)/k^{3/2}}{1/k^{3/2}} = \lim_{k \to \infty} \ln(k) \). Since this limit diverges to infinity, refine the approach by noting that \( \ln(k) \) grows slower than any power of \( k \), so the original series behaves similarly to the convergent p-series, indicating convergence.

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주요 개념

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Convergence of Infinite Series

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely. Determining convergence involves analyzing the behavior of the terms and applying appropriate tests to see if the series sums to a finite value.

Comparison Test

The Comparison Test involves comparing the given series to a known benchmark series with positive terms. If the given series' terms are smaller than those of a convergent series, it also converges; if larger than a divergent series, it diverges. This test is useful when terms resemble simpler series.

Behavior of Logarithmic and Power Functions in Series

Understanding how logarithmic functions like ln(k) grow compared to power functions like k^(3/2) is crucial. Since ln(k) grows slower than any positive power of k, the term ln(k)/k^(3/2) behaves similarly to 1/k^(3/2) for large k, which helps in applying convergence tests effectively.

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.∑ (from k = 3 to ∞)ln(k) / k³ᐟ²