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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.3.85
Chapter 10, Problem 10.3.85

72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 1 to ∞) (((1/6)ᵏ + (1/3)ᵏ) × k⁻¹)

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Identify the given series: \( \sum_{k=1}^{\infty} \left( \left( \frac{1}{6} \right)^k + \left( \frac{1}{3} \right)^k \right) \times \frac{1}{k} \). This is a sum of two series combined inside the summation.
Rewrite the series by separating the sum into two separate series: \( \sum_{k=1}^{\infty} \frac{\left( \frac{1}{6} \right)^k}{k} + \sum_{k=1}^{\infty} \frac{\left( \frac{1}{3} \right)^k}{k} \). This allows us to analyze each series individually.
Recognize that each series is of the form \( \sum_{k=1}^{\infty} \frac{x^k}{k} \), which is related to the Taylor series expansion of the function \( -\ln(1 - x) \) for \( |x| < 1 \).
Check the convergence criteria: since \( \left| \frac{1}{6} \right| < 1 \) and \( \left| \frac{1}{3} \right| < 1 \), both series converge absolutely.
Express each series using the logarithmic function: \( \sum_{k=1}^{\infty} \frac{x^k}{k} = -\ln(1 - x) \). Substitute \( x = \frac{1}{6} \) and \( x = \frac{1}{3} \) respectively, then add the two results to write the sum in closed form.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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 is essential before evaluating the sum, often using tests like the comparison, ratio, or root tests.
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Convergence of an Infinite Series

p-Series and Geometric Series

A p-series has terms of the form 1/k^p and converges if p > 1. A geometric series has terms of the form ar^k and converges if |r| < 1. The given series combines these forms, so understanding their convergence criteria helps analyze the overall behavior.
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Term-by-Term Analysis and Limit Comparison Test

When a series is a sum of multiple sequences, analyzing each part separately can clarify convergence. The Limit Comparison Test compares terms of the given series to a known benchmark series to determine convergence or divergence, especially useful for series with combined terms.
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Related Practice
Textbook Question

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.


∑ (k = 1 to ∞) ((3k³ + 4)(7k² + 1)) / ((2k³ + 1)(4k³ − 1))

Textbook Question

13–20. Explicit formulas Write the first four terms of the sequence { aₙ }∞ₙ₌₁. 

aₙ = 1 + sin(πn / 2)

Textbook Question

Find a formula for the nth partial sum Sₙ of

∑ k = 1 to ∞[(1/(k + 3)) − (1/(k + 4))]

Use your formula to find the sum of the first 36 terms of the series.

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Textbook Question

40–62. Choose your test Use the test of your choice to determine whether the following series converge.

∑ (k = 1 to ∞) k⁸ / (k¹¹ + 3)

Textbook Question

21–42. Geometric series Evaluate each geometric series or state that it diverges.  


37.1 + e/π + e²/π² + e³/π³ + ⋯

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Textbook Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) (2k + 1)! / (k!)²