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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
Capitolo 10, Problema 10.R.89c

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

Guida verificata passo dopo passo
1
Identify the series given: \( \sum_{k=1}^{\infty} \frac{1}{k^5} \). This is a p-series with \( p = 5 \), which converges because \( p > 1 \).
Recognize that the partial sum \( S_n = \sum_{k=1}^n \frac{1}{k^5} \) approximates the total sum, and the remainder (or tail) \( R_n = \sum_{k=n+1}^\infty \frac{1}{k^5} \) represents the error between \( S_n \) and the exact sum.
Use the integral test remainder estimates to find bounds for the remainder \( R_n \). The integral test tells us that for a decreasing positive function \( f(k) = \frac{1}{k^5} \), the remainder satisfies: \[ \int_{n+1}^\infty f(x) \, dx \leq R_n \leq \int_n^\infty f(x) \, dx \] which translates to: \[ \int_{n+1}^\infty \frac{1}{x^5} \, dx \leq R_n \leq \int_n^\infty \frac{1}{x^5} \, dx \]
Calculate these improper integrals (without evaluating the final numeric value) using the formula for integrals of power functions: \[ \int_a^\infty x^{-p} \, dx = \frac{a^{-(p-1)}}{p-1} \quad \text{for} \quad p > 1 \] Apply this to \( p = 5 \) and \( a = n \) and \( a = n+1 \) to express the bounds for \( R_n \).
Finally, express the lower and upper bounds for the exact sum of the series as: \[ L_n = S_n + \int_{n+1}^\infty \frac{1}{x^5} \, dx \quad \text{and} \quad U_n = S_n + \int_n^\infty \frac{1}{x^5} \, dx \] where \( S_n = \sum_{k=1}^n \frac{1}{k^5} \) is the partial sum up to \( n = 5 \).

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Convergent Infinite Series

A convergent infinite series is a sum of infinitely many terms that approaches a finite limit as more terms are added. For example, the p-series ∑ 1/k^p converges if p > 1. Understanding convergence ensures the series has a well-defined sum to approximate.
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Partial Sums and Remainder Estimation

Partial sums are the sums of the first n terms of a series and serve as approximations to the total sum. The remainder (or error) is the difference between the exact sum and the partial sum. Estimating this remainder helps find bounds for the series' exact value.
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Integral Test for Bounding Series Remainders

The integral test compares a series to an improper integral to determine convergence and estimate remainders. For decreasing positive terms, the remainder after n terms is bounded by integrals of the function from n to infinity, providing lower and upper bounds for the series sum.
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{Use of Tech} A savings plan

James begins a savings plan in which he deposits \$100 at the beginning of each month into an account that earns 9% interest annually, or equivalently, 0.75% per month.

To be clear, on the first day of each month, the bank adds 0.75% of the current balance as interest, and then James deposits \$100.


Let Bₙ be the balance in the account after the nᵗʰ payment, where B₀ = \$0.


a.Write the first five terms of the sequence {Bₙ}.

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Explain why or why not

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

a.If limₙ→∞aₙ = 1 and limₙ→∞bₙ = 3, then limₙ→∞(bₙ / aₙ) = 3.

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

∑ (from k = 0 to ∞)(tan⁻¹(k + 2) − tan⁻¹k)

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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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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)

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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 = 3 to ∞)ln(k) / k³ᐟ²

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