Domanda del libro di testo
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]
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Ch. 10 - Infinite Sequences and Series
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 10, Problema 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) ]
Guida verificata passo dopo passo1
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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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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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 term signs. Absolute convergence implies stronger convergence and often simplifies analysis of power series.
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Percorso guidato
07:51Choosing 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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Percorso guidato
07:51Choosing a Convergence Test
Pratica correlata
Domanda del libro di testo
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
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Domanda del libro di testo
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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Domanda del libro di testo
In Exercises 37–42, find the series’ radius of convergence.
∑ (from n = 1 to ∞) [ n! xⁿ / nⁿ ]
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Domanda del libro di testo
Use series to evaluate the limits in Exercises 29–40.
37. lim (x → 0) ln(1 + x²) / (1 - cos(x))
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Domanda del libro di testo
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 = 0 to ∞) [ n xⁿ / (4ⁿ (n² + 1)) ]
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