11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞)1 / ln(eᵏ + 1)

검증된 단계별 안내

First, write down the general term of the series: \(a_k = \frac{1}{\ln(e^k + 1)}\).

Simplify the expression inside the logarithm for large \(k\). Since \(e^k\) grows very fast, \(e^k + 1 \approx e^k\), so \(\ln(e^k + 1) \approx \ln(e^k) = k\).

Using this approximation, the general term behaves like \(a_k \approx \frac{1}{k}\) for large \(k\).

Recall that the harmonic series \(\sum \frac{1}{k}\) diverges, so by the Comparison Test or Limit Comparison Test, compare \(a_k\) with \(\frac{1}{k}\) to determine convergence.

Calculate the limit \(\lim_{k \to \infty} \frac{a_k}{1/k}\) to apply the Limit Comparison Test and conclude whether the original series converges or diverges.

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주요 개념

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Convergence of Infinite Series

An infinite series converges if the sequence of its partial sums approaches a finite limit. Determining convergence involves analyzing the behavior of the terms and applying appropriate tests to see if the sum settles to a finite value or diverges.

Comparison Test

The Comparison Test involves comparing the given series to a known benchmark series. If the terms of the given series are smaller than those of a convergent series, it converges; if larger than those of a divergent series, it diverges. This test helps in establishing convergence by bounding.

Behavior of Logarithmic Functions in Series

Understanding how logarithmic functions grow is crucial when they appear in series terms. Since ln(e^k + 1) behaves roughly like k for large k, the terms 1/ln(e^k + 1) behave like 1/k, which is a harmonic-type term influencing convergence analysis.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.∑ (from k = 1 to ∞)1 / ln(eᵏ + 1)