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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 10 - Infinite Sequences and Series문제 10.PE.47
10장, 문제 10.PE.47

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 ∞) (x + 4)ⁿ/(n3ⁿ)

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1
Identify the given power series: \(\displaystyle \sum_{n=1}^{\infty} \frac{(x+4)^n}{n 3^n}\). Notice that this is a power series centered at \(x = -4\) because the term is \((x+4)^n\).
To find the radius of convergence, apply the Root Test or Ratio Test. The Ratio Test is often easier here. Consider the general term \(a_n = \frac{(x+4)^n}{n 3^n}\). Compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Calculate the ratio inside the limit: \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x+4)^{n+1}}{(n+1) 3^{n+1}} \cdot \frac{n 3^n}{(x+4)^n} \right| = \left| \frac{(x+4)}{3} \cdot \frac{n}{n+1} \right|\). As \(n \to \infty\), \(\frac{n}{n+1} \to 1\), so the limit simplifies to \(L = \left| \frac{x+4}{3} \right|\).
The Ratio Test tells us the series converges if \(L < 1\), so the radius of convergence \(R\) satisfies \(\left| x+4 \right| < 3\). This means the radius of convergence is \(R = 3\), and the interval of convergence is \((-4 - 3, -4 + 3)\) or \((-7, -1)\), but we still need to check the endpoints \(x = -7\) and \(x = -1\) separately.
To determine convergence at the endpoints, substitute \(x = -7\) and \(x = -1\) into the series and analyze the resulting series. For each endpoint, check if the series converges absolutely (by testing the absolute value of terms) or conditionally (converges but not absolutely). This often involves comparing to known series like the p-series or alternating series tests.

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주요 개념

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Radius and Interval of Convergence

The radius of convergence is the distance from the center of a power series within which the series converges. The interval of convergence includes all x-values for which the series converges, possibly including endpoints. Finding these involves applying tests like the Ratio or Root Test to determine where the series converges absolutely.
추천 영상:
07:36
Radius of Convergence

Absolute Convergence

A series converges absolutely if the series of absolute values converges. This means ∑|a_n| converges, ensuring the original series converges regardless of sign changes. Absolute convergence guarantees stronger convergence properties and is often easier to test using comparison or ratio tests.
추천 영상:
가이드 코스
07:51
Choosing a Convergence Test

Conditional Convergence

Conditional convergence occurs when a series converges, but does not converge absolutely. This means the series ∑a_n converges, but ∑|a_n| diverges. Identifying conditional convergence often requires testing endpoints of the interval of convergence separately using tests like the Alternating Series Test.
추천 영상:
가이드 코스
07:51
Choosing a Convergence Test
관련 실천
교과서 질문

Theory and Examples

Suppose that a₁, a₂, a₃, …, aₙ are positive numbers satisfying the following conditions:

i) a₁ ≥ a₂ ≥ a₃ ≥ …;


ii) the series a₂ + a₄ + a₈ + a₁₆ + … diverges.

Show that the series


a₁/1 + a₂/2 + a₃/3 + …


diverges.

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교과서 질문

Convergent Series

Find the sums of the series in Exercises 19–24.


∑ (from n = 2 to ∞) -2/[n(n+1)]

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교과서 질문

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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교과서 질문

Determining Convergence of Series

Which of the series in Exercises 25–44 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers.


∑ (from n = 2 to ∞) 1/[n(ln n)²]

교과서 질문

30. b. By differentiating the series in part (a) term by term, show that

Σ(from n=1 to ∞) n / (n + 1)! = 1.

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교과서 질문

Maclaurin Series

Find Taylor series at x = 0 for the functions in Exercises 63–70.

cos (x³/√5)

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