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Chapter 3: Problem 4

A power series with positive radius of convergence \(r<\infty\) converges absolutely either for all points or for no point on the boundary of the convergence domain. Give examples for these cases.

Short Answer

Expert verified
A power series may either converge or diverge at the boundary of its radius of convergence, depending on the series.

Step by step solution

01

Understanding the Problem

We need to explore how a power series behaves at the boundary of its convergence domain. A power series converges absolutely inside its radius of convergence and might converge or not on the boundary depending on the specific series.
02

Example 1: Converging on the Boundary

Consider the power series \( \sum_{n=1}^{\infty} \frac{1}{n^2} x^n \). The radius of convergence is 1. At the boundary point \( x = 1 \), the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges (since it's a p-series with \( p=2 > 1 \)), hence converges absolutely at the boundary point.
03

Example 2: Not Converging on the Boundary

Consider another series, \( \sum_{n=1}^{\infty} x^n \). The radius of convergence is also 1. At the boundary point \( x = 1 \), the series becomes \( \sum_{n=1}^{\infty} 1 \), which diverges. Thus, the series does not converge absolutely at the boundary.

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Key Concepts

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

Radius of Convergence
When working with power series, the radius of convergence is a critical concept that tells us where a series behaves well. A power series such as \( \sum_{n=0}^{\infty} a_n x^n \) has a radius of convergence \( r \), which determines the interval around zero where the series converges. This radius is found using the formula \( r = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \). The series converges absolutely for all \( |x| < r \). This radius can sometimes turn out to be infinite, implying the series converges everywhere or can be zero, suggesting convergence only at the origin.
The radius of convergence provides a boundary within which we can be certain of convergence. Beyond the radius, behavior becomes unpredictable, necessitating further analysis. Each point on this boundary, often called the "circle of convergence," might show different behaviors, prompting the need to separately analyze every boundary point.
Absolute Convergence
Absolute convergence in power series refers to the situation where not only does the series converge, but it does so in a manner that is unaffected by rearrangement of its terms. Specifically, a series \( \sum_{n=0}^{\infty} a_n x^n \) converges absolutely at point \( x \) if the series \( \sum_{n=0}^{\infty} |a_n x^n| \) converges. For a power series within its radius of convergence (i.e., \( |x| < r \)), absolute convergence is guaranteed. Inside this interval, no rearrangement of terms can lead to divergence, making absolute convergence a particularly strong form of convergence.
Absolute convergence is key in many mathematical analyses due to its stability and applicability in functions' various operations like integration and differentiation, as these processes require convergent series to maintain validity across transformations.
Boundary Behavior
Examining the behavior of a power series at its boundary is a subtle but crucial part of understanding its full scope. While a series converges absolutely within its radius of convergence, the scenario at the boundary can vary widely. Take, for example, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} x^n \). When \( x = 1 \), it converges as a p-series with \( p=2 > 1 \). Here, the boundary supports convergence.
In contrast, the series \( \sum_{n=1}^{\infty} x^n \) has the same radius of convergence \( r = 1 \). At \( x = 1 \), it turns into the harmonic series \( \sum_{n=1}^{\infty} 1 \) which diverges.
This unpredictable behavior on the boundary underlines the importance of testing each boundary point individually to understand if the series maintains convergence or veers into divergence. Thus, even when the radius is known, full comprehension requires examining these individual boundary cases.

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Most popular questions from this chapter

Let \(u\) be a non-constant harmonic function on a domain \(D \subset \mathbb{R}^{2}\). Show that \(u(D)\) is an open interval.

Show: $$ \int_{0}^{\infty} \frac{\log ^{2} x}{1+x^{2}} d x=\frac{\pi^{3}}{8}, \quad \int_{0}^{\infty} \frac{\log x}{1+x^{2}} d x=0. $$

Let \(\sum_{n=0}^{\infty} a_{n} z^{n}\) be a power series with radius of convergence \(r\) Show: (a) If \(R:=\lim _{n \rightarrow \infty}\left|a_{n}\right| /\left|a_{n+1}\right|\) exists, then \(r=R\). (b) If \(\tilde{\rho} \lim _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|} \in[0, \infty]\) exists, then \(r=1 / \tilde{\rho} .\) Here we formally use the conventions \(1 / 0=\infty\) and \(1 / \infty=0 .(r=0\) for \(\tilde{\rho}=\infty\), and \(r=\infty\) for \(\tilde{\rho}=0 .)\) (c) If we set $$ \rho:=\lim _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|}:=\lim _{n \rightarrow \infty}\left(\sup \left\\{\sqrt[n]{\left|a_{n}\right|}, \sqrt[n+1]{\left|a_{n+1}\right|}, \sqrt[n+2]{\left|a_{n+1}\right|}, \ldots\right\\}\right), $$ then - following the same conventions as in (b)- one hasa $$ r=1 / \rho $$

The power series \(\sum_{n=0}^{\infty} c_{n} z^{n}\) and the termwise differentiated power series \(\sum_{n=1}^{\infty} n c_{n} z^{n-1}\) have the same radius of convergence \(r\). Moreover, for all \(z \in U_{r}(0)\) one has \(P^{\prime}(z)=Q(z)\) Hint. For \(z, b \in U_{r}(0)\) $$ \begin{gathered} P(z)-P(b)=\sum_{n=0}^{\infty} c_{n}\left(z^{n}-b^{n}\right)=(z-b) \sum_{n=1}^{\infty} c_{n} \varphi_{n}(z) \\ \text { with } \varphi_{n}(z)=z^{n-1}+z^{n-2} b+\cdots+z b^{n-2}+b^{n-1}. \end{gathered} $$

Let \(f\) be an analytic function on an open set containing the closed disk \(\bar{U}_{r}(a)\). We assume \(|f(a)|<|f(z)|\) for all \(z\) on the boundary of the disk. Then there exists a zero of \(f\) in the interior of the disk. Using this, find a further proof of the Open Mapping Theorem.
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