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
Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.8.71
Capitolo 10, Problema 10.8.71

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


∑ (from k = 1 to ∞)(ln²k) / k³ᐟ²

Guida verificata passo dopo passo
1
Identify the general term of the series: \(a_k = \frac{(\ln k)^2}{k^{3/2}}\).
Recall that for series with positive terms, common convergence tests include the Comparison Test, the Limit Comparison Test, and the p-series test.
Note that the denominator \(k^{3/2}\) is a power function with exponent \(\frac{3}{2} > 1\), which suggests the series resembles a p-series \(\sum \frac{1}{k^p}\) with \(p = \frac{3}{2}\).
Since \(\ln^2 k\) grows slower than any positive power of \(k\), compare \(a_k\) to \(\frac{1}{k^{3/2}}\) using the Limit Comparison Test: compute \(\lim_{k \to \infty} \frac{a_k}{1/k^{3/2}} = \lim_{k \to \infty} (\ln k)^2\).
Because \(\lim_{k \to \infty} (\ln k)^2 = \infty\), the Limit Comparison Test with \(1/k^{3/2}\) is inconclusive; instead, use the Comparison Test noting that for large \(k\), \((\ln k)^2\) grows slower than any power \(k^{\epsilon}\) for \(\epsilon > 0\), so \(a_k\) behaves like \(\frac{1}{k^{3/2 - \epsilon}}\) which still converges since \(3/2 - \epsilon > 1\). Therefore, the series converges by comparison to a convergent p-series.

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:
2m

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 as the index grows large, ensuring the sum does not diverge to infinity.
Video consigliato:
Percorso guidato
06:52
Convergence of an Infinite Series

Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence. The Comparison Test uses inequalities, while the Limit Comparison Test uses the limit of the ratio of terms to decide if both series share the same convergence behavior.
Video consigliato:
Percorso guidato
07:45
Limit Comparison Test

Behavior of Logarithmic and Power Functions in Series

Understanding how logarithmic terms like (ln k)² grow relative to polynomial terms like k^(3/2) is crucial. Since polynomial terms grow faster than logarithmic terms, this relationship helps in applying comparison tests to determine series convergence.
Video consigliato:
07:32
Representing Functions as Power Series
Pratica correlata
Domanda del libro di testo

11–27. Alternating Series Test Determine whether the following series converge.

∑ (k = 1 to ∞) (−1)ᵏ⁺¹ k² / (k³ + 1)

62
views
Domanda del libro di testo

Property of divergent series Prove Property 2 of Theorem 10.8: If ∑ aₖ diverges, then ∑ caₖ also diverges, for any real number c ≠ 0.

69
views
Domanda del libro di testo

55–70. More sequences

Find the limit of the following sequences or determine that the sequence diverges.


{nsin³(nπ / 2) / (n + 1)}"

47
views
Domanda del libro di testo

9–30. The Ratio and Root Tests Use the Ratio Test or the Root Test to determine whether the following series converge absolutely or diverge.

1 + (1 / 2)² + (1 / 3)³ + (1 / 4)⁴ + ⋯

69
views
Domanda del libro di testo

9–16. Divergence Test Use the Divergence Test to determine whether the following series diverge or state that the test is inconclusive.

∑ (k = 2 to ∞) k / ln k

84
views
Domanda del libro di testo

What comparison series would you use with the Comparison Test to determine whether ∑ (k = 1 to ∞) 2ᵏ / (3ᵏ + 1) converges?

66
views