Problem 6
Question
Suppose \(\sum a_{n} x^{n}\) converges for \(x=1\). a) What can you say about the radius of convergence? b) If you further know that at \(x=1\) the convergence is not absolute, what can you say?
Step-by-Step Solution
Verified Answer
a) The radius of convergence \(R\) is at least 1. b) \(R=1\) since the series converges conditionally at \(x=1\).
1Step 1: Understanding the Problem
We are given that the power series \(\sum a_{n} x^{n}\) converges for \(x=1\). This means that when we replace \(x\) with 1 in the series, it converges to a finite sum.
2Step 2: Applying the Definition of Radius of Convergence
The radius of convergence \(R\) of a power series \(\sum a_n x^n\) is defined as the distance from the center of the series (typically 0) to the boundary of the interval within which the series converges.
3Step 3: Determining Radius from Given Convergence at a Point
Since the series converges at \(x=1\), by definition, \(|1| = 1\) is included in the interval of convergence. Therefore, \(R \geq 1\).
4Step 4: Analyzing Convergence Types for Further Information
We're given that the convergence at \(x=1\) is not absolute, meaning \(\sum |a_n|\) diverges. This specific point of non-absolute convergence usually implies that \(x=1\) is precisely at the radius boundary, suggesting \(R=1\).
Key Concepts
Power SeriesAbsolute ConvergenceInterval of Convergence
Power Series
A power series is a series of the form \( \sum a_{n} x^{n} \), where each term is a multiple of an increasing power of a variable \( x \). The coefficients \( a_n \) are constant terms, and \( n \) is a non-negative integer.
Power series can resemble polynomials but with infinitely many terms. However, unlike polynomials, these series can lead to divergent or convergent summations depending on the value of \( x \).
Power series play a crucial role in calculus and analysis because they can represent a function as an infinite sum of terms, providing insights into the behavior of the function within a specific range.
Power series can resemble polynomials but with infinitely many terms. However, unlike polynomials, these series can lead to divergent or convergent summations depending on the value of \( x \).
Power series play a crucial role in calculus and analysis because they can represent a function as an infinite sum of terms, providing insights into the behavior of the function within a specific range.
- For example, the well-known exponential function \( e^x \) can be expressed as the power series \( \sum \frac{x^n}{n!} \).
- Power series help in approximation problems and solving differential equations.
- Each power series is centered around a specific point, typically \( x = 0 \), unless specified otherwise."
Absolute Convergence
Absolute convergence refers to a condition in which not only does a series converge, but the series formed by taking the absolute values of its terms also converges. For a power series \( \sum a_n x^n \), absolute convergence at a point \( x = a \) means that \( \sum |a_n| |a|^n \) converges.
This is an important distinction for power series as absolute convergence signifies a stronger form of convergence. When a series converges absolutely, any reordering of its terms will still result in convergence to the same sum.
This is an important distinction for power series as absolute convergence signifies a stronger form of convergence. When a series converges absolutely, any reordering of its terms will still result in convergence to the same sum.
- If a series converges absolutely at a certain \( x \), then it will converge at all smaller \( x \) values in magnitude as well.
- Absolute convergence leads to many useful properties, like allowing you to use algebraic manipulations safely on series.
Interval of Convergence
The interval of convergence of a power series \( \sum a_n x^n \) is the range of \( x \) values for which the series converges to a finite sum. The concept relies heavily on the radius of convergence, \( R \), which defines the radius of the interval around the center where the series is assured to converge.
Within this interval, the series will converge for successive values of \( x \). The boundary points, however, need specifics checks as the series may converge or diverge at these edges.
For our example, since we know it converges at \( x = 1 \) and this convergence is not absolute, \( x = 1 \) likely sits on the fringe of the convergence interval.
Within this interval, the series will converge for successive values of \( x \). The boundary points, however, need specifics checks as the series may converge or diverge at these edges.
For our example, since we know it converges at \( x = 1 \) and this convergence is not absolute, \( x = 1 \) likely sits on the fringe of the convergence interval.
- The typical form of the interval, if the series is centered at \( 0 \), will be \( (-R, R) \).
- At the boundaries, \( x = -R \) and \( x = R \), further investigation is needed to determine convergence.
- In our study here, the non-absolute convergence hints that the radius of convergence might exactly be 1, making the interval \( [-1, 1] \) or \( (-1, 1] \) depending on boundary behavior.
Other exercises in this chapter
Problem 5
For \(j=1,2, \ldots, n\), let \(\left\\{x_{j k}\right\\}_{k=1}^{\infty}\) denote n sequences. Suppose that for each \(j\) $$ \sum_{k=1}^{\infty} x_{j, k} $$ is
View solution Problem 5
In the following exercises, feel free to use what you know from calculus to find the limit, if it exists. But you must prove that you found the correct limit, o
View solution Problem 6
Suppose \(\left|x_{n}-x_{k}\right| \leq n / k^{2}\) for all \(n\) and \(k .\) Show that \(\left\\{x_{n}\right\\}\) is Cauchy.
View solution Problem 6
Let \(\left\\{x_{n}\right\\}\) and \(\left\\{y_{n}\right\\}\) be bounded sequences such that \(x_{n} \leq y_{n}\) for all \(n .\) Then show that $$ \limsup _{n
View solution