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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.4.39a
Capitolo 10, Problema 10.4.39a

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.


a. Use Sₙ to estimate the sum of the series.


39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Guida verificata passo dopo passo
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, which provides bounds for the remainder \( R_n = S - S_n \) of a convergent series with positive, decreasing terms. The theorem states that the remainder is bounded by the integral test inequalities:
\[ \int_{n+1}^{\infty} f(x) \, dx \leq R_n \leq \int_n^{\infty} f(x) \, dx, \]
where \( f(x) = \frac{1}{x^7} \) in this problem. To estimate the sum \( S \), first compute the partial sum \( S_n = \sum_{k=1}^n \frac{1}{k^7} \) for \( n = 2 \).
Next, calculate the integrals to find the bounds for the remainder:
\[ \int_3^{\infty} \frac{1}{x^7} \, dx \quad \text{and} \quad \int_2^{\infty} \frac{1}{x^7} \, dx. \]
Use these integral values to find the lower and upper bounds for the total sum \( S \) by adding them to \( S_2 \):
\[ S_2 + \int_3^{\infty} \frac{1}{x^7} \, dx \leq S \leq S_2 + \int_2^{\infty} \frac{1}{x^7} \, dx. \]

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Convergent Series

A convergent series is an infinite sum whose partial sums approach a finite limit. For example, the series ∑ 1/k^7 converges because the terms decrease rapidly and satisfy the p-series test with p = 7 > 1. Understanding convergence ensures that the sum S exists and can be approximated.
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Convergence of an Infinite Series

Partial Sums (Sₙ)

The partial sum Sₙ is the sum of the first n terms of a series and serves as an approximation to the total sum. For the series ∑ 1/k^7, S₂ = 1 + 1/2^7 provides an estimate of the infinite sum. Partial sums are fundamental in estimating series values and analyzing error bounds.
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Integration Using Partial Fractions

Theorem 10.13 (Bounds on Remainder of a Series)

Theorem 10.13 gives lower and upper bounds for the remainder (error) when approximating a convergent series by its partial sum Sₙ. It typically uses integrals to bound the tail of the series, allowing us to estimate how close Sₙ is to the actual sum. This theorem is essential for quantifying the accuracy of partial sums.
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Alternating Series Remainder
Pratica correlata
Domanda del libro di testo

{Use of Tech} Drug Dosing

A patient takes 75 mg of a medication every 12 hours; 60% of the medication in the blood is eliminated every 12 hours.



a.Let dₙ equal the amount of medication (in mg) in the bloodstream after n doses, where d₁ = 75.

Find a recurrence relation for dₙ.

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

a. Find the next two terms of the sequence.


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

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

a. Suppose 0 < aₖ < bₖ. If ∑ (k = 1 to ∞) aₖ converges, then ∑ (k = 1 to ∞) bₖ converges.

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Explain why or why not

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


a.The sequence of partial sums for the series1 + 2 + 3 + ⋯ is {1, 3, 6, 10, …}.

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72–75. {Use of Tech} Practical sequences

Consider the following situations that generate a sequence


a.Write out the first five terms of the sequence.


Radioactive decay

A material transmutes 50% of its mass to another element every 10 years due to radioactive decay. Let Mₙ be the mass of the radioactive material at the end of the nᵗʰ decade, where the initial mass of the material is M₀ = 20g.

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


a. n!n! = (2n)! for all positive integers n.

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