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

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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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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.

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Convergence of an Infinite Series

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.

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Direct Comparison Test

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.

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