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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.5.65b
Chapter 10, Problem 10.5.65b

Assume that bₙ is a sequence of positive numbers converging to 4/5. Determine if the following series converge or diverge.
b. ∑ (from n = 1 to ∞) (5/4)ⁿ (bₙ)

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Identify the general term of the series: \(a_n = \left(\frac{5}{4}\right)^n b_n\), where \(b_n\) is a sequence converging to \(\frac{4}{5}\).
Since \(b_n \to \frac{4}{5}\) as \(n \to \infty\), analyze the behavior of \(a_n\) by considering the limit \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(\frac{5}{4}\right)^n b_n\).
Because \(b_n\) approaches a positive constant \(\frac{4}{5}\), the dominant factor in \(a_n\) is \(\left(\frac{5}{4}\right)^n\), which grows exponentially since \(\frac{5}{4} > 1\).
Recall that for a series \(\sum a_n\) to converge, the terms \(a_n\) must approach zero as \(n\) approaches infinity. Check if \(\lim_{n \to \infty} a_n = 0\).
Since \(\left(\frac{5}{4}\right)^n\) grows without bound and \(b_n\) approaches a positive constant, \(a_n\) does not approach zero. Therefore, by the Divergence Test (also called the nth-term test), the series \(\sum_{n=1}^\infty \left(\frac{5}{4}\right)^n b_n\) diverges.

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Key Concepts

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

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence approach as n becomes very large. In this problem, knowing that bₙ converges to 4/5 helps us understand the behavior of the terms in the series for large n, which is crucial for analyzing convergence.
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Introduction to Sequences

Geometric Series and Its Convergence

A geometric series has terms of the form arⁿ, where r is the common ratio. Such a series converges if and only if |r| < 1. Here, the factor (5/4)ⁿ suggests a geometric component with ratio greater than 1, which typically leads to divergence unless modified by other factors.
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Comparison Test for Series Convergence

The comparison test involves comparing a given series to a known benchmark series to determine convergence or divergence. Since bₙ approaches 4/5, the terms behave like (5/4)ⁿ times a positive constant, allowing us to compare with a geometric series to conclude about the series' behavior.
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Related Practice
Textbook Question

b. From Example 5, Section 10.2, show that

S = 1 + ∑(from n=1 to ∞) [1 / (n²(n + 1))].

Textbook Question

A sequence of rational numbers is described as follows:

1/1,3/2,7/5,17/12,…,a/b,(a + 2b)/(a + b),…

Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let xₙ and yₙ be, respectively, the numerator and the denominator of the nᵗʰ fraction rₙ = xₙ / yₙ.

b. The fractions rₙ = xₙ / yₙ approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that rₙ² − 2 = ±(1 / yₙ)² and that yₙ is not less than n.)

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Textbook Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (b) absolutely?

∑ (from n = 1 to ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]

Textbook Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (c) conditionally?

∑ (from n = 1 to ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]

Textbook Question

Intervals of Convergence

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (b) absolutely?

∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]

Textbook Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (b) absolutely?

∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]