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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.89b
Chapter 10, Problem 10.R.89b

89–90. {Use of Tech} Lower and upper bounds of a series
For each convergent series and given value of n, complete the following.


b. Find an upper bound for the remainder Rₙ.


89.∑ (from k = 1 to ∞)1 / k⁵ ;n = 5

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1
Identify the series given: \( \sum_{k=1}^{\infty} \frac{1}{k^5} \). This is a p-series with \( p = 5 \), which converges because \( p > 1 \).
Recall that the remainder \( R_n \) after \( n \) terms is the difference between the infinite sum and the partial sum up to \( n \): \( R_n = S - S_n \). We want to find an upper bound for \( R_n \) when \( n = 5 \).
Use the integral test remainder estimate for a decreasing positive function \( f(k) = \frac{1}{k^5} \). The upper bound for the remainder \( R_n \) is given by the integral from \( n \) to infinity of \( f(x) \): \[ R_n \leq \int_n^{\infty} \frac{1}{x^5} \, dx \]
Evaluate the improper integral: \[ \int_n^{\infty} \frac{1}{x^5} \, dx = \lim_{t \to \infty} \int_n^t x^{-5} \, dx \] Find the antiderivative of \( x^{-5} \), which is \( \frac{x^{-4}}{-4} = -\frac{1}{4x^4} \).
Apply the limits to the antiderivative: \[ \lim_{t \to \infty} \left( -\frac{1}{4t^4} + \frac{1}{4n^4} \right) = 0 + \frac{1}{4n^4} = \frac{1}{4n^4} \] Thus, the upper bound for the remainder \( R_n \) when \( n = 5 \) is \( \frac{1}{4 \times 5^4} \).

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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/k⁵ converges because its terms decrease rapidly and resemble a p-series with p > 1. Understanding convergence ensures that the remainder after n terms is meaningful and finite.
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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 sum. 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

Integral Test and Remainder Bounds

The integral test can be used to estimate the remainder of a decreasing positive term series. It states that the remainder Rₙ is less than or equal to the integral of the function from n to infinity. This provides a practical way to find upper bounds for the error in series approximations.
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Integral Test
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

Textbook Question

Explain why or why not

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


g.The series ∑ (from k = 1 to ∞) (k² / (k² + 1)) converges.

Textbook Question

27–37. Evaluating series Evaluate the following infinite series or state that the series diverges.

∑ (from k = 0 to ∞)((1/3)ᵏ + (4/3)ᵏ)

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.


a.How far does the crew dig in 10 weeks? 20 weeks? N weeks?

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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 k = 1 to ∞)k! / (eᵏ kᵏ)

Textbook Question

89–90. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, complete the following.


c. Find lower and upper bounds (Lₙ and Uₙ respectively) for the exact value of the series.


89.∑ (from k = 1 to ∞)1 / k⁵ ;n = 5