Textbook Question
Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence.
∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]
Ch. 10 - Infinite Sequences and Series
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 10, Problem 10.7.28a
Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence.
∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]
Verified step by step guidance1
Identify the general term of the series: \(a_n = (-2)^n (n + 1) (x - 1)^n\).
Apply the Ratio Test to find the radius of convergence. Compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Substitute \(a_{n+1}\) and \(a_n\) into the ratio: \(\left| \frac{(-2)^{n+1} (n+2) (x-1)^{n+1}}{(-2)^n (n+1) (x-1)^n} \right| = \left| -2 \cdot \frac{n+2}{n+1} \cdot (x-1) \right|\).
Simplify the expression inside the limit: \(L = \lim_{n \to \infty} 2 \cdot \frac{n+2}{n+1} \cdot |x-1| = 2 |x-1|\).
Set the limit \(L < 1\) for convergence, which gives \(2 |x-1| < 1\). Solve this inequality to find the radius of convergence and the interval centered at \(x=1\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Radius of Convergence
The radius of convergence is the distance from the center of a power series within which the series converges absolutely. It can be found using the Ratio Test or Root Test on the general term of the series. This radius defines an interval around the center where the series behaves well.
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07:36
Radius of Convergence
Interval of Convergence
The interval of convergence is the set of all x-values for which the power series converges. It is centered at the series’ expansion point and extends to the radius of convergence on both sides. Endpoints must be checked separately to determine if the series converges or diverges there.
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08:44Interval of Convergence
Ratio Test for Series Convergence
The Ratio Test determines convergence by examining 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 to find the radius of convergence.
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05:35Ratio Test
Related Practice
Textbook Question
Power Series
In Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally.
∑ (from n = 1 to ∞) xⁿ/nⁿ
Textbook Question
Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series.
a. Converges absolutely for x = 1
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
The series
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ⋯
converges to eˣ for all x.
a. Find a series for (d/dx)eˣ. Do you get the series for eˣ? Explain your answer.