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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 10.4.39b
Chapter 10, Problem 10.4.39b

39–40. {Use of Tech} Lower and upper bounds of a series
For each convergent series and given value of n, use Theorem 10.13 to complete the following.


b. Find an upper bound for the remainder Rₙ.


39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Verified step by step guidance
1
Identify the series given: \( \sum_{k=1}^{\infty} \frac{1}{k^7} \). This is a p-series with \( p = 7 \), which converges because \( p > 1 \).
Recall Theorem 10.13 (the Integral Test remainder estimate), which states that for a decreasing, positive, continuous function \( f(k) \), the remainder \( R_n = S - S_n \) satisfies the inequality \( R_n \leq \int_{n}^{\infty} f(x) \, dx \).
Set \( f(x) = \frac{1}{x^7} \) and write the upper bound for the remainder as \( R_n \leq \int_{n}^{\infty} \frac{1}{x^7} \, dx \).
Evaluate the improper integral \( \int_{n}^{\infty} x^{-7} \, dx \) by finding the antiderivative of \( x^{-7} \), which is \( \frac{x^{-6}}{-6} \), and then compute the limit as the upper bound approaches infinity.
Substitute \( n = 2 \) into the evaluated integral to express the upper bound for the remainder \( R_2 \) explicitly in terms of \( n \).

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Key Concepts

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

Convergent Series

A convergent series is an infinite sum whose partial sums approach a finite limit. For example, the series ∑ 1/k⁷ converges because the terms decrease rapidly and satisfy the p-series test with p = 7 > 1. Understanding convergence ensures that the remainder or error after n terms is well-defined.
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Convergence of an Infinite Series

Remainder (Error) in a Series Approximation

The remainder Rₙ is the difference between the infinite series sum and the partial sum up to n terms. It measures the error when approximating the series by a finite number of terms. Finding an upper bound for Rₙ helps estimate how close the partial sum is to the actual sum.
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Alternating Series Remainder

Theorem 10.13 (Integral Test Remainder Estimate)

Theorem 10.13 provides bounds for the remainder of a convergent series whose terms come from a positive, decreasing function. It states that the remainder Rₙ is less than or equal to the integral of the function from n to infinity. This theorem allows calculation of explicit upper bounds for the error.
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Integral Test
Related Practice
Textbook Question

Explain why or why not

Determine whether the following statements are true and give an explanation or counterexample.



b.If limₙ→∞aₙ = 0 and limₙ→∞bₙ = ∞, then limₙ→∞aₙbₙ = 0.

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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.


41. ∑ (k = 1 to ∞) 1 / k⁶

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)?

Textbook Question

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


c. Suppose f is a continuous, positive, decreasing function, for x ≥ 1, and aₖ = f(k), for k = 1, 2, 3, …. If ∑ (k = 1 to ∞) aₖ converges to L, then ∫ (1 to ∞) f(x) dx converges to L.

Textbook Question

57–60. Heights of bouncing balls A ball is thrown upward to a height of hₒ meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let hₙ be the height after the nth bounce. Consider the following values of hₒ and r.


b. Find an explicit formula for the nth term of the sequence {hₙ}.


h₀ = 30,r = 0.25

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