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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
Chapter 10, Problem 10.1.67b

67–70. Formulas for sequences of partial sums Consider the following infinite series.


b.Find a formula for the nth partial sum Sₙ of the infinite series. Use this formula to find the next four partial sums S₅, S₆, S₇, S₈ of the infinite series.


∑⁽∞⁾ₖ₌₁2⁄[(2k − 1)(2k + 1)]

Verified step by step guidance
1
Start by writing the nth partial sum of the series as the sum of the first n terms: \(S_n = \sum_{k=1}^n \frac{2}{(2k - 1)(2k + 1)}\).
Use partial fraction decomposition to rewrite the general term \(\frac{2}{(2k - 1)(2k + 1)}\) in a form that will allow telescoping. Set \(\frac{2}{(2k - 1)(2k + 1)} = \frac{A}{2k - 1} + \frac{B}{2k + 1}\) and solve for A and B.
After finding A and B, rewrite each term of the sum using these partial fractions. This should create a telescoping series where most terms cancel out when summed.
Express \(S_n\) as a simplified expression after cancellation, which will be a formula involving only a few terms depending on n.
Use the formula for \(S_n\) to calculate the next four partial sums: \(S_5\), \(S_6\), \(S_7\), and \(S_8\) by substituting n = 5, 6, 7, and 8 respectively.

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Key Concepts

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

Partial Sums of a Series

A partial sum Sₙ is the sum of the first n terms of a series. It helps approximate the value of an infinite series by adding a finite number of terms. Understanding partial sums is essential to analyze convergence and to find explicit formulas for sums up to a certain index.
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Intro to Series: Partial Sums

Telescoping Series

A telescoping series is one where many terms cancel out when the partial sums are expanded. This simplification often allows finding a closed-form expression for Sₙ. Recognizing telescoping patterns is key to simplifying complex series like the given one.
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Geometric Series

Decomposition into Partial Fractions

Partial fraction decomposition breaks a complex rational expression into simpler fractions. This technique is useful to rewrite terms in a series to reveal telescoping behavior. Applying it to the given term helps in finding a formula for the nth partial sum.
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Partial Fraction Decomposition: Distinct Linear Factors
Related Practice
Textbook Question

41–44. {Use of Tech} Remainders and estimates Consider the following convergent series.


b. Find how many terms are needed to ensure that the remainder is less than 10⁻³.


43. ∑ (k = 1 to ∞) 1 / 3ᵏ

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Textbook Question

27–34. Working with sequences Several terms of a sequence {aₙ}ₙ₌₁∞ are given.

b. Find a recurrence relation that generates the sequence (supply the initial value of the index and the first term of the sequence).


{1, 3, 9, 27, 81, ......}

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Textbook Question

Suppose the sequence {aₙ}⁽∞⁾ₙ₌₀ is defined by the recurrence relation

aₙ₊₁ = ⅓aₙ + 6;a₀ = 3.


b.Explain why {aₙ}⁽∞⁾ₙ₌₀ converges and find the limit.

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Textbook Question

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.

b. Which of the following series converges faster? Explain.

∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?

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Textbook Question

72–75. {Use of Tech} Practical sequences

Consider the following situations that generate a sequence


b.Find an explicit formula for the terms of the sequence.


Drug elimination

Jack took a 200-mg dose of a pain killer at midnight. Every hour, 5% of the drug is washed out of his bloodstream. Let dₙ be the amount of drug in Jack’s blood n hours after the drug was taken, where d₀ = 200mg.

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. A series that converges absolutely must converge.

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