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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 10.3.83
Chapter 10, Problem 10.3.83

72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 1 to ∞) ((1/3) × (5/6)ᵏ + (3/5) × (7/9)ᵏ)

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Recognize that the given series is a sum of two separate infinite series: \( \sum_{k=1}^\infty \left( \frac{1}{3} \times \left( \frac{5}{6} \right)^k + \frac{3}{5} \times \left( \frac{7}{9} \right)^k \right) = \sum_{k=1}^\infty \frac{1}{3} \left( \frac{5}{6} \right)^k + \sum_{k=1}^\infty \frac{3}{5} \left( \frac{7}{9} \right)^k \).
Identify each series as a geometric series of the form \( \sum_{k=1}^\infty ar^k \), where \(a\) is the coefficient and \(r\) is the common ratio. For the first series, \(a = \frac{1}{3}\) and \(r = \frac{5}{6}\). For the second series, \(a = \frac{3}{5}\) and \(r = \frac{7}{9}\).
Check the convergence of each geometric series by verifying if the absolute value of the common ratio \( |r| < 1 \). Since \( \frac{5}{6} < 1 \) and \( \frac{7}{9} < 1 \), both series converge.
Use the formula for the sum of an infinite geometric series starting at \(k=1\): \[ S = \frac{ar}{1 - r} \]. Apply this formula to each series separately.
Calculate the sum of the original series by adding the sums of the two geometric series: \[ S_{total} = \frac{\frac{1}{3} \times \frac{5}{6}}{1 - \frac{5}{6}} + \frac{\frac{3}{5} \times \frac{7}{9}}{1 - \frac{7}{9}} \].

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Key Concepts

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

Infinite Geometric Series

An infinite geometric series is a sum of terms where each term is a constant multiple (common ratio) of the previous one. It converges if the absolute value of the common ratio is less than 1, and its sum can be found using the formula S = a / (1 - r), where a is the first term and r is the common ratio.
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Convergence and Divergence of Series

A series converges if the sequence of its partial sums approaches a finite limit; otherwise, it diverges. For geometric series, convergence depends on the common ratio's magnitude. Understanding convergence is essential to determine whether the series sum exists or not.
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Convergence of an Infinite Series

Linearity of Series Summation

The sum of a series that is a sum of two series equals the sum of each series separately. This property allows breaking down complex series into simpler parts, evaluating each independently, and then combining the results to find the total sum.
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Related Practice
Textbook Question

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


∑ (from k = 0 to ∞)k² · 1.001⁻ᵏ

Textbook Question

33–38. {Use of Tech} Remainders in alternating series Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10⁻⁴ in magnitude. Although you do not need it, the exact value of the series is given in each case.


π / 4 = ∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

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Textbook Question

55–70. More sequences

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


{(75n⁻¹ / 99ⁿ) + (5ⁿsinn / 8ⁿ)}

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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 ∞) 1 / (2k − √k)

Textbook Question

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.

∑ (from k = 1 to ∞) ((-1)ᵏ⁺¹ × k²ᵏ) / (k! × k!)

Textbook Question

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


∑ (k = 0 to ∞) k / (2k + 1)