Domanda del libro di testo
b. From Example 5, Section 10.2, show that
S = 1 + ∑(from n=1 to ∞) [1 / (n²(n + 1))].
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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 10.7.32b
Intervals of Convergence
Intervals of Convergence
In Exercises 1–36, for what values of x does the series converge (b) absolutely?
∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]
Guida verificata passo dopo passo1
Identify the given series: \( \sum_{n=1}^{\infty} \frac{(3x + 1)^{n+1}}{2n + 2} \). We want to find the values of \( x \) for which this series converges absolutely.
To check for absolute convergence, consider the series of absolute values: \( \sum_{n=1}^{\infty} \left| \frac{(3x + 1)^{n+1}}{2n + 2} \right| = \sum_{n=1}^{\infty} \frac{|3x + 1|^{n+1}}{2n + 2} \).
Since the denominator \( 2n + 2 \) grows linearly and does not affect the radius of convergence, focus on the term \( |3x + 1|^{n+1} \). Use the root test or ratio test to find the radius of convergence with respect to \( |3x + 1| \).
Apply the root test: compute \( \lim_{n \to \infty} \sqrt[n]{\frac{|3x + 1|^{n+1}}{2n + 2}} = \lim_{n \to \infty} \frac{|3x + 1|^{(n+1)/n}}{(2n + 2)^{1/n}} = |3x + 1| \), since the other terms tend to 1.
Set the root test limit less than 1 for absolute convergence: \( |3x + 1| < 1 \). Solve this inequality for \( x \) to find the interval of absolute convergence.
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Interval of Convergence
The interval of convergence is the set of all x-values for which a given power series converges. It is found by determining where the series converges absolutely or conditionally, often using tests like the Ratio or Root Test. This interval can be finite or infinite and includes checking endpoints separately.
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08:44
Interval of Convergence
Absolute Convergence
A series converges absolutely if the series of absolute values converges. Absolute convergence implies convergence regardless of the sign of terms, making it a stronger form of convergence. For power series, testing absolute convergence helps identify the radius and interval of convergence.
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Percorso guidato
07:51Choosing a Convergence Test
Ratio Test
The Ratio Test determines convergence by examining the limit of the ratio of consecutive terms' absolute values. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges. It is especially useful for power series to find the radius of convergence.
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Percorso guidato
05:35Ratio Test
Pratica correlata
Domanda del libro di testo
Assume that bₙ is a sequence of positive numbers converging to 4/5. Determine if the following series converge or diverge.
b. ∑ (from n = 1 to ∞) (5/4)ⁿ (bₙ)
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Domanda del libro di testo
A sequence of rational numbers is described as follows:
1/1,3/2,7/5,17/12,…,a/b,(a + 2b)/(a + b),…
Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let xₙ and yₙ be, respectively, the numerator and the denominator of the nᵗʰ fraction rₙ = xₙ / yₙ.
b. The fractions rₙ = xₙ / yₙ approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that rₙ² − 2 = ±(1 / yₙ)² and that yₙ is not less than n.)
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Domanda del libro di testo
Intervals of Convergence
In Exercises 1–36, for what values of x does the series converge (b) absolutely?
∑ (from n = 1 to ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]
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Domanda del libro di testo
Intervals of Convergence
In Exercises 1–36, for what values of x does the series converge (c) conditionally?
∑ (from n = 1 to ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]
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Domanda del libro di testo
Use the Cauchy condensation test from Exercise 59 to show that:
b. ∑ (from n=1 to ∞) [1 / nᵖ] converges if p > 1 and diverges if p ≤ 1.
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