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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 10.7.13
Chapter 10, Problem 10.7.13

9–30. The Ratio and Root Tests Use the Ratio Test or the Root Test to determine whether the following series converge absolutely or diverge.
∑ (from k = 1 to ∞) (k² / 4ᵏ)

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Identify the series given: \( \sum_{k=1}^{\infty} \frac{k^{2}}{4^{k}} \). We want to determine if it converges absolutely or diverges.
Recall the Ratio Test: For a series \( \sum a_k \), compute the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). If \( L < 1 \), the series converges absolutely; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.
Set \( a_k = \frac{k^{2}}{4^{k}} \). Compute \( \frac{a_{k+1}}{a_k} = \frac{(k+1)^{2} / 4^{k+1}}{k^{2} / 4^{k}} = \frac{(k+1)^{2}}{k^{2}} \cdot \frac{1}{4} \).
Take the limit as \( k \to \infty \) of \( \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \frac{(k+1)^{2}}{k^{2}} \cdot \frac{1}{4} = \left( \lim_{k \to \infty} \frac{(k+1)^{2}}{k^{2}} \right) \cdot \frac{1}{4} \).
Evaluate the limit \( \lim_{k \to \infty} \frac{(k+1)^{2}}{k^{2}} = 1 \), so the overall limit \( L = 1 \cdot \frac{1}{4} = \frac{1}{4} \). Since \( L < 1 \), by the Ratio Test, the series converges absolutely.

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Key Concepts

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

Ratio Test

The Ratio Test determines the convergence of a series by examining the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive.
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Ratio Test

Root Test

The Root Test analyzes the nth root of the absolute value of the nth term of a series. If the limit of this root is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive. It is especially useful for series with terms raised to the nth power.
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Absolute Convergence

A series converges absolutely if the series of the absolute values of its terms converges. Absolute convergence guarantees convergence of the original series and is a stronger condition than conditional convergence, ensuring stability under rearrangement of terms.
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Choosing a Convergence Test
Related Practice
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^(–1/5)

Textbook Question

9–15. Geometric sums Evaluate each geometric sum.


{Use of Tech}∑ k = 0 to 20(2/5)²ᵏ

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Textbook Question

Growth rates of sequences

Use Theorem 10.6 to find the limit of the following sequences or state that they diverge.


aₙ = (6ⁿ + 3ⁿ) / (6ⁿ + n¹⁰⁰)

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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

Textbook Question

39–44. {Use of Tech} Estimating infinite series Estimate the value of the following convergent series with an absolute error less than 10⁻³.


∑ (k = 1 to ∞) (−1)ᵏ / k⁵

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Textbook Question

17–22. Integral Test Use the Integral Test to determine whether the following series converge after showing that the conditions of the Integral Test are satisfied.

∑ (k = 1 to ∞) 1 / (∛(5k + 3))