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
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 10.7.32a
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) ]
Verified step by step guidance1
Identify the general term of the series: \( a_n = \frac{(3x + 1)^{n+1}}{2n + 2} \).
To find the radius of convergence, apply the Root Test or Ratio Test. Here, the Ratio Test is convenient. Consider \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
Compute the ratio \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(3x + 1)^{n+2}}{2(n+1) + 2} \cdot \frac{2n + 2}{(3x + 1)^{n+1}} \right| = \left| (3x + 1) \right| \cdot \frac{2n + 2}{2n + 4} \).
Take the limit as \( n \to \infty \) of the ratio: \( \lim_{n \to \infty} \left| (3x + 1) \right| \cdot \frac{2n + 2}{2n + 4} = \left| 3x + 1 \right| \cdot 1 = \left| 3x + 1 \right| \).
Set the limit less than 1 for convergence: \( \left| 3x + 1 \right| < 1 \). Solve this inequality to find the interval of convergence, then check the endpoints \( x \) values by substituting back into the original series to determine if the series converges at those points.
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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 on the x-axis where the series behaves well and sums to a finite value.
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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 tested separately to determine if the series converges or diverges there.
Recommended video:
08:44Interval of Convergence
Applying the Ratio Test to Power Series
The Ratio Test compares the limit of the absolute value of consecutive terms to determine convergence. For power series, it helps find the radius of convergence by simplifying the ratio of terms involving x. If the limit is less than one, the series converges; if greater, it diverges.
Recommended video:
05:58Intro to Power Series
Related Practice
Textbook Question
Determining Convergence of Sequences
Which of the sequences whose nth terms appear in Exercises 1–18 converge, and which diverge? Find the limit of each convergent sequence.
aₙ = 1 + (0.9)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
Determining Convergence of Sequences
Which of the sequences whose nth terms appear in Exercises 1–18 converge, and which diverge? Find the limit of each convergent sequence.
aₙ = (-4)ⁿ/n!
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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.
Textbook Question
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 ]