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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 10.6.63
Chapter 10, Problem 10.6.63

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (k = 1 to ∞) (−1)ᵏ⁺¹ · (k!) / (kᵏ)(Hint: Show that k! / kᵏ ≤ 2 / k², for k ≥ 3.)

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First, identify the general term of the series: \[a_k = (-1)^{k+1} \cdot \frac{k!}{k^k}\]. Since the series is alternating due to the factor \[(-1)^{k+1}\], we consider both absolute and conditional convergence.
To check for absolute convergence, examine the absolute value of the terms: \[|a_k| = \frac{k!}{k^k}\]. We want to determine if the series \[\sum_{k=1}^\infty \frac{k!}{k^k}\] converges.
Use the given hint to compare \[\frac{k!}{k^k}\] with \[\frac{2}{k^2}\] for \[k \geq 3\]. Since \[\frac{k!}{k^k} \leq \frac{2}{k^2}\], and the series \[\sum \frac{2}{k^2}\] converges (p-series with \[p=2 > 1\]), by the Comparison Test, the series of absolute values converges.
Since the series of absolute values converges, the original alternating series converges absolutely. Absolute convergence implies convergence regardless of the alternating sign.
Therefore, conclude that the series \[\sum_{k=1}^\infty (-1)^{k+1} \frac{k!}{k^k}\] converges absolutely.

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Key Concepts

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

Absolute and Conditional Convergence

A series converges absolutely if the series of absolute values converges. If the original series converges but not absolutely, it is conditionally convergent. Understanding this distinction helps determine the nature of convergence for alternating or complex series.
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Choosing a Convergence Test

Comparison Test

The comparison test involves comparing a given series to a known benchmark series to determine convergence. If the terms of the given series are smaller than those of a convergent series, it also converges. This test is useful when bounding terms like k!/k^k by simpler expressions.
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Direct Comparison Test

Factorials and Exponential Growth

Factorials (k!) grow faster than polynomial terms but slower than some exponentials. The term k!/k^k involves factorial growth divided by exponential growth, and analyzing their relative rates is key to estimating term size and applying convergence tests effectively.
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Factorials
Related Practice
Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{tan⁻¹(n)}

Textbook Question

23–38. Divergence, Integral, and p-series Tests Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge.

∑ (k = 1 to ∞) (k / (k + 10))ᵏ

Textbook Question

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) (−1)ᵏ / k^(2/3)

Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{ⁿ√(e³ⁿ⁺⁴)} 

1
views
Textbook Question

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.


∑ (k = 1 to ∞) sin(1 / k) / k²

Textbook Question

The terms of a sequence of partial sums are defined by Sₙ = ∑ⁿₖ₌₁ k² , for n=1, 2, 3, .....Evaluate the first four terms of the sequence.