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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.R.31
Chapter 10, Problem 10.R.31

27–37. Evaluating series Evaluate the following infinite series or state that the series diverges.
∑ (from k = 1 to ∞)ln((2k + 1) / (2k − 1))

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1
Start by writing out the general term of the series: \( a_k = \ln\left(\frac{2k + 1}{2k - 1}\right) \).
Use the logarithm property \( \ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B) \) to rewrite the term as \( a_k = \ln(2k + 1) - \ln(2k - 1) \).
Express the partial sum \( S_n = \sum_{k=1}^n a_k \) by substituting the expanded terms: \( S_n = \sum_{k=1}^n \left( \ln(2k + 1) - \ln(2k - 1) \right) \).
Rewrite the partial sum as \( S_n = \left( \ln 3 - \ln 1 \right) + \left( \ln 5 - \ln 3 \right) + \left( \ln 7 - \ln 5 \right) + \cdots + \left( \ln(2n + 1) - \ln(2n - 1) \right) \) and observe the telescoping pattern where most terms cancel out.
Simplify the telescoping sum to \( S_n = \ln(2n + 1) - \ln 1 \), then analyze the limit \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \ln(2n + 1) \) to determine whether the series converges or diverges.

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Key Concepts

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

Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. To evaluate such a series, it is crucial to determine whether it converges (approaches a finite limit) or diverges (grows without bound or oscillates). Convergence tests help decide if the sum exists.
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Convergence of an Infinite Series

Properties of Logarithms

Logarithmic properties, such as ln(a/b) = ln(a) - ln(b), allow simplification of terms in the series. Applying these properties can transform the series into a telescoping form or other manageable expressions, facilitating evaluation.
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Change of Base Property

Telescoping Series

A telescoping series is one where many terms cancel out when the series is expanded, leaving only a few terms to sum. Recognizing telescoping behavior is key to simplifying and finding the sum of certain infinite series involving differences of logarithms.
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Geometric Series
Related Practice
Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)(1 − cos(1 / k))²

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42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)tanh(k)

Textbook Question

Sequences versus series

a.Find the limit of the sequence { (−⁴⁄₅)ᵏ }.

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Textbook Question

25–26. Recursively defined sequences

The following sequences {aₙ} from n = 0 to ∞ are defined by a recurrence relation. Assume each sequence is monotonic and bounded.


b.Determine the limit of each sequence.


25.aₙ₊₁ = (1 / 2) aₙ + 8;a₀ = 80

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Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from j = 0 to ∞)2 ⋅ 4ʲ / (2j + 1)!

Textbook Question

Building a tunnel — first scenario

A crew of workers is constructing a tunnel through a mountain. Understandably, the rate of construction decreases because rocks and earth must be removed a greater distance as the tunnel gets longer. Suppose each week the crew digs 0.95 of the distance it dug the previous week. In the first week, the crew constructed 100 m of tunnel.


b.What is the longest tunnel the crew can build at this rate?