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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
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.7.14
Chapter 10, Problem 10.7.14

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 ∞) [ (x − 1)ⁿ / (n³ 3ⁿ) ]

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1
Identify the given power series: \(\sum_{n=1}^{\infty} \frac{(x - 1)^n}{n^3 3^n}\).
To find the radius of convergence, apply the Root Test or Ratio Test. Here, the Ratio Test is convenient. Consider the general term \(a_n = \frac{(x - 1)^n}{n^3 3^n}\) and compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Calculate the ratio: \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x - 1)^{n+1}}{(n+1)^3 3^{n+1}} \cdot \frac{n^3 3^n}{(x - 1)^n} \right| = \left| \frac{x - 1}{3} \right| \cdot \frac{n^3}{(n+1)^3}\).
Evaluate the limit as \(n \to \infty\): \(L = \left| \frac{x - 1}{3} \right| \cdot \lim_{n \to \infty} \frac{n^3}{(n+1)^3} = \left| \frac{x - 1}{3} \right|\).
The Ratio Test states the series converges if \(L < 1\), so the radius of convergence \(R\) satisfies \(\left| x - 1 \right| < 3\). This gives the interval \((1 - 3, 1 + 3)\) or \((-2, 4)\). Next, check convergence at the endpoints \(x = -2\) and \(x = 4\) by substituting back into the series and analyzing absolute and conditional 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 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.
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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 term signs. Absolute convergence implies stronger convergence and often simplifies analysis of power series.
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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.
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Choosing a Convergence Test
Related Practice
Textbook Question

Determining Convergence or Divergence

In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑(from n=1 to ∞) [(-1)ⁿ n² e⁻ⁿ]

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) = √(1 − x),a = 0

Textbook Question

Determining Convergence or Divergence

Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.

∑ (from n=1 to ∞) (1 − n) / n2ⁿ

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Textbook Question

In Exercises 37–42, find the series’ radius of convergence.

∑ (from n = 1 to ∞) [ n! xⁿ / nⁿ ]

Textbook Question

Use series to evaluate the limits in Exercises 29–40.

37. lim (x → 0) ln(1 + x²) / (1 - cos(x))

Textbook Question

In Exercises 67–72, use the results of Exercises 63 and 64 to determine if each series converges or diverges.

∑(from n=2 to ∞) [(ln n)¹⁰⁰⁰ / n¹.⁰⁰¹]

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