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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 10.6.25
Chapter 10, Problem 10.6.25

11–27. Alternating Series Test Determine whether the following series converge.
∑ (k = 1 to ∞) (−1)ᵏ⁺¹ k¹/ᵏ

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Identify the general term of the series: \(a_k = (-1)^{k+1} k^{1/k}\). This is an alternating series because of the factor \((-1)^{k+1}\), which causes the terms to alternate in sign.
Recall the Alternating Series Test (Leibniz Test), which states that an alternating series \(\sum (-1)^k b_k\) converges if two conditions are met: (1) the sequence \(b_k\) is positive, decreasing, and (2) \(\lim_{k \to \infty} b_k = 0\).
Set \(b_k = k^{1/k}\) (the absolute value of the terms without the alternating sign). Check if \(b_k\) is positive and decreasing for sufficiently large \(k\). Since \(k^{1/k} > 0\) for all \(k\), the positivity condition is satisfied.
Analyze the limit \(\lim_{k \to \infty} k^{1/k}\). Use the fact that \(k^{1/k} = e^{(\ln k)/k}\). Evaluate the limit of the exponent \(\lim_{k \to \infty} \frac{\ln k}{k}\), which approaches 0, so the limit of \(k^{1/k}\) approaches \(e^0 = 1\).
Since \(\lim_{k \to \infty} b_k = 1 \neq 0\), the Alternating Series Test fails, and therefore the series does not converge by this test.

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Key Concepts

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

Alternating Series Test

The Alternating Series Test determines the convergence of series whose terms alternate in sign. It requires that the absolute value of the terms decreases monotonically to zero. If these conditions hold, the series converges, even if it does not converge absolutely.
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Alternating Series Test

Limit of the General Term

For any infinite series to converge, the limit of its general term as k approaches infinity must be zero. If the terms do not approach zero, the series diverges. This is a necessary condition for convergence but not sufficient on its own.
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One-Sided Limits

Behavior of the Term k^(1/k)

The term k^(1/k) represents the k-th root of k, which approaches 1 as k becomes very large. Understanding this limit helps determine whether the terms of the series decrease to zero, a key step in applying the Alternating Series Test.
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Divergence Test (nth Term Test)
Related Practice
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 ∞) ((3k³ + 4)(7k² + 1)) / ((2k³ + 1)(4k³ − 1))

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Textbook Question

13–52. Limits of sequences

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


{bₙ}, where

bₙ = { n / (n + 1)if n ≤ 5000

ne⁻ⁿif n > 5000 }

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Textbook Question

21–42. Geometric series Evaluate each geometric series or state that it diverges.  


29.∑ (k = 1 to ∞) e^(–2k)

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Textbook Question

9–16. Divergence Test Use the Divergence Test to determine whether the following series diverge or state that the test is inconclusive.

∑ (k = 0 to ∞) 1 / (1000 + k)

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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 ∞) (2 + (−1)ᵏ) / k²

36
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Textbook Question

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

{Use of Tech} ∑ (k = 1 to ∞) 1 / (4lnk)

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