72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))

검증된 단계별 안내

First, rewrite the general term of the series to simplify the expression inside the summation. The term is given by \(\frac{\ln\left((k+1)k^{-1}\right)}{\ln k \times \ln(k+1)}\). Use the logarithm property \(\ln(a/b) = \ln a - \ln b\) to rewrite the numerator as \(\ln(k+1) - \ln k\).

Substitute the simplified numerator back into the term to get \(\frac{\ln(k+1) - \ln k}{\ln k \times \ln(k+1)}\). Then, separate this fraction into two parts: \(\frac{\ln(k+1)}{\ln k \times \ln(k+1)} - \frac{\ln k}{\ln k \times \ln(k+1)}\).

Simplify each part of the separated fraction. The first part simplifies to \(\frac{1}{\ln k}\) and the second part simplifies to \(\frac{1}{\ln(k+1)}\). So the general term becomes \(\frac{1}{\ln k} - \frac{1}{\ln(k+1)}\).

Recognize that the series is telescoping because each term is of the form \(a_k - a_{k+1}\), where \(a_k = \frac{1}{\ln k}\). Write out the first few terms explicitly to see the cancellation pattern.

Use the telescoping property to express the partial sum \(S_n = \sum_{k=2}^n \left( \frac{1}{\ln k} - \frac{1}{\ln(k+1)} \right)\) as \(\frac{1}{\ln 2} - \frac{1}{\ln(n+1)}\). Then analyze the limit of \(S_n\) as \(n \to \infty\) to determine whether the series converges or diverges.

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주요 개념

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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. Determining convergence often involves applying tests that analyze the behavior of the terms or partial sums.

Properties of Logarithms

Logarithmic properties, such as ln(a/b) = ln(a) - ln(b) and ln(ab) = ln(a) + ln(b), help simplify complex expressions. Recognizing these can transform the series terms into simpler forms, making it easier to analyze or compare with known series.

Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence. The limit comparison test uses the limit of the ratio of terms from two series; if the limit is finite and positive, both series share the same convergence behavior.

72–86. Evaluating series Evaluate each series or state that it diverges.∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))