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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
Chapter 10, Problem 10.7.17

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 ∞) ((-7)ᵏ / k²)

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Identify the series given: \( \sum_{k=1}^{\infty} \frac{(-7)^k}{k^2} \). We want to determine if it converges absolutely or diverges.
Recall that absolute convergence means the series \( \sum_{k=1}^{\infty} \left| a_k \right| \) converges, where \( a_k = \frac{(-7)^k}{k^2} \). So consider the series \( \sum_{k=1}^{\infty} \frac{7^k}{k^2} \).
Apply the Ratio Test, which uses the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). For our terms, calculate \( \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(-7)^{k+1} / (k+1)^2}{(-7)^k / k^2} \right| = \frac{7^{k+1}}{7^k} \cdot \frac{k^2}{(k+1)^2} = 7 \cdot \left( \frac{k}{k+1} \right)^2 \).
Evaluate the limit as \( k \to \infty \): \( L = 7 \cdot \lim_{k \to \infty} \left( \frac{k}{k+1} \right)^2 = 7 \cdot 1 = 7 \).
Interpret the result of the Ratio Test: since \( L = 7 > 1 \), the series \( \sum_{k=1}^{\infty} \frac{7^k}{k^2} \) diverges, so the original series does not converge absolutely. Because the terms do not tend to zero in absolute value, the original series diverges.

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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 terms in a series. If the limit of this nth root as n approaches infinity 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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Absolute Convergence

A series converges absolutely if the series of the absolute values of its terms converges. Absolute convergence guarantees convergence regardless of the sign of terms, which is important when applying tests like the Ratio or Root Test to series with alternating or negative terms.
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Choosing a Convergence Test
Related Practice
Textbook Question

Suppose the sequence { aₙ} is defined by the explicit formula aₙ = 1/n, for n=1, 2, 3, .....Write out the first five terms of the sequence.

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Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.


∑ (from k = 1 to ∞)(cos(1 / k) – cos(1 / (k + 1)))

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Textbook Question

43–44. Periodic doses

Suppose you take a dose of m mg of a particular medication once per day. Assume f equals the fraction of the medication that remains in your blood one day later. Just after taking another dose of medication on the second day, the amount of medication in your blood equals the sum of the second dose and the fraction of the first dose remaining in your blood, which is m + mf. Continuing in this fashion, the amount of medication in your blood just after your nth dose is


Aₙ = m + mf + ⋯ + mfⁿ⁻¹.


For the given values of f and m, calculate A₅, A₁₀, A₃₀, and lim (n → ∞) Aₙ. Interpret the meaning of the limit lim (n → ∞) Aₙ.


43.f = 0.25,m = 200 mg

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Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞) 2⁹k / kᵏ

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Textbook Question

What test is advisable if a series involves a factorial term?

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Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.


∑ (from k = 1 to ∞)(1 / √(k + 2) – 1 / √k)

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