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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.67
Chapter 10, Problem 10.R.67

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)k⁵ e⁻ᵏ

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Identify the series given: \( \sum_{k=1}^{\infty} k^{5} e^{-k} \). This is an infinite series where the general term is \( a_k = k^{5} e^{-k} \).
Recognize that the term \( e^{-k} \) can be rewritten as \( \left( \frac{1}{e} \right)^k \), which is an exponential decay factor.
Consider using the Ratio Test for convergence, which is effective for series involving factorials, exponentials, or powers. The Ratio Test states to compute \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
Calculate the ratio \( \frac{a_{k+1}}{a_k} = \frac{(k+1)^5 e^{-(k+1)}}{k^5 e^{-k}} = \frac{(k+1)^5}{k^5} \cdot e^{-1} \). Simplify this expression to prepare for taking the limit as \( k \to \infty \).
Evaluate the limit \( L = \lim_{k \to \infty} \frac{(k+1)^5}{k^5} \cdot e^{-1} \). Since \( \frac{(k+1)^5}{k^5} \to 1 \), the limit simplifies to \( L = e^{-1} \). Use the Ratio Test conclusion: if \( L < 1 \), the series converges; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.

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Key Concepts

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

Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. Determining whether such a series converges means checking if the sum approaches a finite limit as the number of terms grows indefinitely. Understanding convergence is essential to analyze the behavior of series like ∑ k⁵ e⁻ᵏ.
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Convergence of an Infinite Series

Comparison and Limit Comparison Tests

These tests help determine convergence by comparing the given series to a known benchmark series. The Comparison Test checks if terms are smaller than those of a convergent series, while the Limit Comparison Test uses the limit of the ratio of terms. They are useful when terms involve products like polynomial and exponential functions.
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Limit Comparison Test

Behavior of Exponential vs. Polynomial Functions

Exponential functions like e⁻ᵏ decay faster than any polynomial grows as k approaches infinity. This means terms like k⁵ e⁻ᵏ tend to zero rapidly, often ensuring convergence of the series. Recognizing this interplay helps in selecting appropriate convergence tests.
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Graphs of Exponential Functions
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 ∞)k⁴ / √(9k¹² + 2)

Textbook Question

Sequences versus series

a.Find the limit of the sequence { (−⁴⁄₅)ᵏ }.

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Textbook Question

25–26. Recursively defined sequences

The following sequences {aₙ} from n = 0 to ∞ are defined by a recurrence relation. Assume each sequence is monotonic and bounded.


b.Determine the limit of each sequence.


25.aₙ₊₁ = (1 / 2) aₙ + 8;a₀ = 80

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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 ∞)3 / (2 + eᵏ)

Textbook Question

77–87. Absolute or conditional convergence

Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞)(−1)ᵏ⁺¹ / k³⁄⁷

Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from j = 0 to ∞)2 ⋅ 4ʲ / (2j + 1)!