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Chapter 9: Problem 34

(a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$\sum_{n=1}^{\infty} \frac{3 \cdot 5 \cdot 7 \cdots(2 n+1)}{n^{2} \cdot 2^{n}} x^{n+1}$$

Short Answer

Expert verified
Radius of convergence: \( R = \frac{1}{2} \). The series converges absolutely for \(-\frac{1}{2} < x < \frac{1}{2}\) and conditionally for no values of \(x\).

Step by step solution

01

Recognize the series format

The given series is \( \sum_{n=1}^{\infty} \frac{3 \cdot 5 \cdot 7 \cdots(2 n+1)}{n^{2} \cdot 2^{n}} x^{n+1} \). It is a power series in terms of \( x \) starting from the term \( x^{n+1} \).
02

Rewrite the series

Rewrite the terms involving \( n\) in the series. Recall that the numerator expands as a product of odd numbers, specifically \( (2n+1)!! = (2n+1)!! = (1)(2)(3)\cdots(2n+1)/(2)(1) \) or \( \frac{(2n+1)!}{2^n \cdot n!} \). Rewrite the entire series as:\[ \sum_{n=1}^{\infty} \frac{(2n+1)!}{(2^n n!) n^2 2^n} x^{n+1} \]
03

Use the Ratio Test to find the Radius of Convergence

For a series \( \sum a_n x^n \), the radius of convergence \( R \) is found by:\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = L \]Here, \( a_n = \frac{(2n+1)!}{n^2 \cdot 2^n} \).Calculate \( \frac{a_{n+1}}{a_n} \):\[ \frac{a_{n+1}}{a_n} = \frac{(2n+3)!}{(n+1)^2 \cdot 2^{n+1}} \cdot \frac{n^2 \cdot 2^n}{(2n+1)!} \]Simplifying gives:\[ \frac{(2n+3)(2n+2)\cdot n^2}{(n+1)^2 \cdot 2} \]As \( n \to \infty \),\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 2 \]Thus, the radius of convergence is \( R = \frac{1}{2} \).
04

Determine the interval of convergence

Given \( R = \frac{1}{2} \), the interval of convergence is:\( -\frac{1}{2} < x < \frac{1}{2} \).
05

Check for absolute convergence

The series \( \sum |a_n x^n| \) converges absolutely if it converges for all \( |x| < \frac{1}{2} \). In this case, the series converges absolutely for \( -\frac{1}{2} < x < \frac{1}{2} \).
06

Check for conditional convergence

Conditional convergence occurs if the series converges when \( |x| = \frac{1}{2} \) but does not converge absolutely. Check the endpoints:1. At \( x = \frac{1}{2} \): series does not converge since the terms do not tend to zero.2. At \( x = -\frac{1}{2} \): series does not converge as these are the same endpoints.Thus, no conditional convergence occurs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a popular method for determining the convergence of a series. It compares the ratio of successive terms in a series to evaluate the convergence behavior.
Here's how it works: Given a series \( \sum a_n \), the Ratio Test considers the limit \( L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.

In the exercise, the Ratio Test helped find the radius of convergence \( R \) by simplifying the ratio of terms. This resulted in \( R = \frac{1}{2} \), which implies that the power series converges in the open interval \(-\frac{1}{2} < x < \frac{1}{2}\). Understanding the Ratio Test is essential for handling power series effectively.
Power Series
A power series is an infinite series of the form \( \sum a_n x^n \), where \( a_n \) are coefficients and \( x \) is a variable. These series resemble polynomial expressions but can extend indefinitely.
Power series converge within a certain range of \( x \) values called the interval of convergence.
The midpoint of this interval is at \( x = 0 \), and the divergence of the series can often be checked using the Ratio Test.
  • Within this interval, the power series behaves like a polynomial, and the concepts of absolute and conditional convergence can be applied.
  • The radius of convergence \( R \) is the distance from the midpoint to the boundary of the interval.

Understanding power series is crucial as they are used to approximate real-world functions and are foundational in calculus.
Absolute Convergence
Absolute convergence refers to a series \( \sum a_n \) where the series formed by taking the absolute value of terms, \( \sum |a_n| \), converges as well. If a series converges absolutely, it means it converges regardless of the order of terms.
For a series to converge absolutely, it must first meet the criteria of convergence for its absolute terms:
  • This nature of convergence ensures stability and guarantees that rearrangements of the series will also converge.
  • In the exercise, the original series converged absolutely for \(-\frac{1}{2} < x < \frac{1}{2}\).

Absolute convergence is a stronger condition than simple convergence. If a series doesn't converge absolutely, it might still converge conditionally.
Conditional Convergence
Conditional convergence occurs when a series \( \sum a_n \) converges, but the series formed by its absolute values, \( \sum |a_n| \), does not converge.
This implies that the series relies heavily on the order of its terms for convergence:
  • A conditionally convergent series is delicate鈥攃hanging the order of the terms can lead to divergence.
  • In the exercise, it was identified that no conditional convergence occurred as the series did not converge at the boundary points \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

Conditional convergence highlights scenarios where convergence is not robust and can be subject to change based on term arrangement.

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Most popular questions from this chapter

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{3^{n}}{n^{n}}$$

Prove that if \(\sum a_{n}\) converges absolutely, then \(\sum a_{n}^{2}\) converges.

Use a geometric series to represent each of the given functions as a power series about \(x=0,\) and find their intervals of convergence. a. \(f(x)=\frac{5}{3-x}\) b. \(g(x)=\frac{3}{x-2}\)

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{n}}{\left(2^{n}\right)^{2}}$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=3}^{\infty} \sin \left(\frac{1}{\sqrt{n}}\right)$$
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