Textbook Question
Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
∑ (from n = 1 to ∞) [ (ln n) xⁿ ]
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 10.10.29
Use series to evaluate the limits in Exercises 29–40.
29. lim (x → 0) (e^x - (1 + x)) / x²
Verified step by step guidance1
Recall the Maclaurin series expansion for the exponential function \(e^x\), which is given by:
\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\]
Substitute the series expansion of \(e^x\) into the given limit expression:
\[\lim_{x \to 0} \frac{e^x - (1 + x)}{x^2} = \lim_{x \to 0} \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) - (1 + x)}{x^2}\]
Simplify the numerator by canceling out the \(1\) and \(x\) terms:
\[\lim_{x \to 0} \frac{\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x^2}\]
Divide each term in the numerator by \(x^2\):
\[\lim_{x \to 0} \left( \frac{1}{2!} + \frac{x}{3!} + \frac{x^2}{4!} + \cdots \right)\]
Evaluate the limit as \(x\) approaches 0 by substituting \(x = 0\) into the simplified expression, which leaves only the constant term(s).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Series Expansion
A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. For functions like e^x, the series expansion around x=0 (Maclaurin series) helps approximate the function using polynomials, which simplifies limit evaluation.
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Limit Evaluation Using Series
By substituting the series expansion into the limit expression, complex functions can be approximated by polynomials. This allows direct simplification and evaluation of limits, especially when the original expression results in indeterminate forms like 0/0.
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Handling Indeterminate Forms
Limits that yield forms like 0/0 require algebraic manipulation or series expansion to resolve. Using series expansions helps rewrite the numerator and denominator to identify the leading terms, enabling the determination of the limit's value.
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Related Practice
Textbook Question
In Exercises 43–50, use Theorem 20 to find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.
∑ (from n = 0 to ∞) [ (x² − 1) / 2 ]ⁿ
Textbook Question
Is it true that a sequence {aₙ} of positive numbers must converge if it is bounded above? Give reasons for your answer.
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Textbook Question
Finding Taylor Polynomials
In Exercises 1–10, find the Taylor polynomials of orders 0, 1, 2, and 3 generated by f at a.
f(x) = sin x,a = 0
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Textbook Question
Finding Taylor Series
Use substitution (as in Formula (7)) to find the Taylor series at x = 0 of the functions in Exercises 1–12.
e⁻ˣ/²
Textbook Question
If ∑aₙ is a convergent series of positive terms, prove that ∑sin(aₙ) converges.
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