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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 10.8.41
Capitolo 10, Problema 10.8.41

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) 2ᵏ / (3ᵏ − 2ᵏ)

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Identify the given series: \( \sum_{k=1}^{\infty} \frac{2^{k}}{3^{k} - 2^{k}} \). We want to determine if this series converges or diverges.
Analyze the general term \( a_k = \frac{2^{k}}{3^{k} - 2^{k}} \). For large \( k \), compare the dominant terms in the denominator to simplify the expression.
Since \( 3^{k} \) grows faster than \( 2^{k} \), for large \( k \), \( 3^{k} - 2^{k} \approx 3^{k} \). So, \( a_k \approx \frac{2^{k}}{3^{k}} = \left( \frac{2}{3} \right)^{k} \).
Use the Comparison Test or Limit Comparison Test by comparing \( a_k \) with the geometric series \( \sum \left( \frac{2}{3} \right)^{k} \), which is a convergent geometric series because \( \left| \frac{2}{3} \right| < 1 \).
Conclude that since \( a_k \) behaves like a convergent geometric series for large \( k \), the original series \( \sum_{k=1}^{\infty} \frac{2^{k}}{3^{k} - 2^{k}} \) converges by the Comparison Test.

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Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. Determining whether such a series converges means checking if the sum approaches a finite limit as the number of terms grows indefinitely. Understanding convergence is essential to analyze the behavior of the given series.
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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 established convergence properties. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than a divergent series, it diverges. This test helps in establishing convergence by bounding the series.
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Direct Comparison Test

Ratio Test

The Ratio Test examines the limit of the ratio of consecutive terms in a series. If this limit is less than one, the series converges absolutely; if greater than one, it diverges. This test is particularly useful for series involving exponential terms, like powers of k, as in the given problem.
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Ratio Test
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Domanda del libro di testo

54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.


59. ∑ (k = –3 to ∞) 4 / ((4k – 3)(4k + 1))

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


∑ (from k = 1 to ∞)(1 + 1 / (2k))ᵏ

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55–70. More sequences

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


{cosn / n}

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51–56. {Use of Tech} Recurrence relations Consider the following recurrence relations. Make a table with at least ten terms and determine a plausible limit of the sequence or state that the sequence diverges.


aₙ₊₁ = 4aₙ + 1 a₀ = 1

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84–87. {Use of Tech} Sequences by recurrence relations

The following sequences, defined by a recurrence relation, are monotonic and bounded, and therefore converge by Theorem 10.5.


a.Examine the first three terms of the sequence to determine whether the sequence is nondecreasing or nonincreasing.

b.Use analytical methods to find the limit of the sequence.


aₙ₊₁ = 2aₙ(1 − aₙ);a₀ = 0.3

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What comparison series would you use with the Limit Comparison Test to determine whether ∑ (k = 1 to ∞) (k² + k + 5) / (k³ + 3k + 1) converges?

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