Skip to main content
Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.8.61

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


∑ (from k = 1 to ∞)1 / ln(eᵏ + 1)

Guida verificata passo dopo passo
1
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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.
Video consigliato:
Percorso guidato
06:52
Convergence of an Infinite Series

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.
Video consigliato:
Percorso guidato
09:25
Direct Comparison Test

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.
Video consigliato:
5:26
Graphs of Logarithmic Functions