Textbook Question
13–20. Explicit formulas Write the first four terms of the sequence { aₙ }∞ₙ₌₁.
aₙ = (−1)ⁿ / 2ⁿ
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Ch. 10 - Sequences and Infinite Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 10.4.27
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^(1/k)
Verified step by step guidance1
First, identify the general term of the series: \(a_k = k^{1/k}\).
Apply the Divergence Test by finding the limit of \(a_k\) as \(k\) approaches infinity: compute \(\lim_{k \to \infty} k^{1/k}\).
Rewrite the term inside the limit using exponentials and logarithms: \(k^{1/k} = e^{(1/k) \ln(k)}\).
Evaluate the limit of the exponent: \(\lim_{k \to \infty} \frac{\ln(k)}{k}\), which approaches 0, so the limit of \(a_k\) is \(e^0 = 1\).
Since the limit of \(a_k\) is 1 (not zero), by the Divergence Test, the series \(\sum_{k=1}^\infty k^{1/k}\) diverges.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Divergence Test
The Divergence Test states that if the limit of the terms of a series does not approach zero as k approaches infinity, then the series diverges. It is a quick initial check to determine if a series can possibly converge.
Recommended video:
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Divergence Test (nth Term Test)
Integral Test
The Integral Test relates the convergence of a series to the convergence of an improper integral of a related function. If the integral of the function from 1 to infinity converges, then the series converges; if the integral diverges, so does the series.
Recommended video:
07:25Integral Test
p-series Test
The p-series Test applies to series of the form ∑ 1/k^p. Such a series converges if and only if p > 1, and diverges otherwise. It is useful for comparing or identifying the behavior of series with terms involving powers of k.
Recommended video:
04:30P-Series and Harmonic Series
Related Practice
Textbook Question
32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞) (2k + 1)! / (k!)²
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Textbook Question
40–62. Choose your test Use the test of your choice to determine whether the following series converge.
∑ (k = 3 to ∞) 1 / lnk
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Textbook Question
11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from j = 1 to ∞) 5 / (j² + 4)
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Textbook Question
54–69. Telescoping series
For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.
61. ∑ (k = 1 to ∞) ln((k + 1) / k)
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Textbook Question
Why does the value of a converging alternating series with terms that are nonincreasing in magnitude lie between any two consecutive terms of its sequence of partial sums?
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