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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
Chapter 10, Problem 10.7.28b

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 ]

Verified step by step guidance
1
Identify the given power series: \(\sum_{n=0}^{\infty} (-2)^n (n+1) (x-1)^n\).
To determine absolute convergence, consider the series formed by the absolute values of the terms: \(\sum_{n=0}^{\infty} |(-2)^n (n+1) (x-1)^n| = \sum_{n=0}^{\infty} (n+1) |2|^n |x-1|^n = \sum_{n=0}^{\infty} (n+1) (2|x-1|)^n\).
Recognize that this is a power series in terms of \(r = 2|x-1|\) with general term \((n+1) r^n\). To analyze convergence, use the root or ratio test, or recall the known result for the series \(\sum (n+1) r^n\).
Recall that the series \(\sum_{n=0}^{\infty} (n+1) r^n\) converges if and only if \(|r| < 1\). Therefore, set up the inequality \(2|x-1| < 1\) to find the interval where the series converges absolutely.
Solve the inequality \(2|x-1| < 1\) to find the values of \(x\) for which the original series converges absolutely. This will give the interval of absolute convergence.

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Key Concepts

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

Radius and Interval of Convergence

The radius of convergence defines the distance from the center point within which a power series converges. The interval of convergence includes all x-values for which the series converges, possibly including endpoints. Finding this interval involves applying tests like the Ratio or Root Test to the general term.
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Radius of Convergence

Absolute Convergence

A series converges absolutely if the series of absolute values converges. This is a stronger form of convergence ensuring the original series converges regardless of term signs. Testing absolute convergence often simplifies analysis by removing alternating signs.
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Choosing a Convergence Test

Ratio Test for Series Convergence

The Ratio Test evaluates the limit of the absolute value of the ratio of consecutive terms. If this limit is less than one, the series converges absolutely; if greater than one, it diverges. This test is especially useful for power series involving factorials or exponential terms.
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Ratio Test
Related Practice
Textbook Question

∑ (from n=1 to ∞) (1 / √(n + 1)) diverges

b. What should n be in order that the partial sum sₙ = ∑ (from i=1 to n) (1 / √(i + 1)) satisfies sₙ > 1000?

Textbook Question

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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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

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function f(x) at x = a is called the quadratic approximation of f at x = a. In Exercises 41–46, find the (a) linearization (Taylor polynomial of order 1)

f(x) = ln(cos x)

Textbook Question

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function f(x) at x = a is called the quadratic approximation of f at x = a. In Exercises 41–46, find the (a) linearization (Taylor polynomial of order 1)

f(x) = 1 / √(1 − x²)

Textbook Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence.

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