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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 10.7.32b
Chapter 10, Problem 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) ]

Verified step by step guidance
1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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Choosing 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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Ratio Test
Related Practice
Textbook Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (c) conditionally?

∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]

Textbook Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (c) conditionally?

∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]

Textbook Question

b. From Example 5, Section 10.2, show that

S = 1 + ∑(from n=1 to ∞) [1 / (n²(n + 1))].

Textbook Question

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 ]

Textbook Question

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 ]

Textbook Question

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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