Problem 32 - A First Course in Complex Analysis with Applications

Question

Show that the power series \(\sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}}\) converges at every point on its circle of convergence.

Step-by-Step Solution

Answer

The series converges at every point on its unit circle.

Step 1: Identify the Radius of Convergence

To determine the radius of convergence of the series \( \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} \), use the root test or ratio test. Here, using the root test:Calculate \[ \lim_{k \to \infty} \sqrt[k]{\left| \frac{z^{k}}{k^{2}} \right|} = \lim_{k \to \infty} \frac{|z|}{k^{2/k}} = |z| \cdot \lim_{k \to \infty} \frac{1}{k^{2/k}} \]Since \( \lim_{k \to \infty} k^{2/k} = 1 \), the limit becomes \( |z| \). Thus, the radius of convergence \( R = 1 \). The series converges for \( |z| < 1 \).

Step 2: Investigate Convergence at Boundary |z| = 1

To establish convergence at the boundary \( |z| = 1 \), substitute \( |z| = 1 \) into the series:\[ \sum_{k=1}^{\infty} \frac{e^{i \theta k}}{k^{2}} \]where \( z = e^{i \theta} \) on the unit circle. The test for convergence of this series will involve the absolute value:\[ \sum_{k=1}^{\infty} \frac{1}{k^{2}} \]This series converges since it is a p-series with \( p = 2 \), which is greater than 1.

Step 3: Conclusion from p-Series

The absolute version \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \) converges by the p-series test (as \( p = 2 > 1 \)). Hence, for any point \( z \) on the unit circle (with \( |z| = 1 \)), the series \( \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} \) converges by the comparison with the convergent p-series.

Key Concepts

Radius of Convergence

The concept of the radius of convergence is essential in determining the values for which a power series converges. For a series \( \sum_{k=0}^{\infty} a_k z^k \), the radius of convergence \( R \) is the distance from the center of the circle within which the series converges. To determine \( R \), we often use the root test or the ratio test. If \( R \) is 1, then the series converges for all values of \( |z| < 1 \).

In our series \( \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} \), the radius of convergence was found to be 1 through the root test. This indicates that the series converges when the magnitude (or absolute value) of \( z \) is less than 1. Points on the circle defined by \( |z| = 1 \) require additional testing to determine convergence.

Root Test

The root test is a powerful tool for finding the radius of convergence of a power series. The test considers the \( k \)-th root of the absolute value of the series' terms as \( k \) approaches infinity. For the series \( \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} \), the expression becomes:

• \( \lim_{k \to \infty} \sqrt[k]{\left| \frac{z^{k}}{k^{2}} \right|} = |z| \cdot \lim_{k \to \infty} \frac{1}{k^{2/k}} \)

As \( k \) becomes very large, \( k^{2/k} \) approaches 1. Thus, the limit simplifies to \( |z| \), and thus the series converges for \( |z| < 1 \).

The root test is particularly useful for series involving power terms like this one, where taking the \( k \)-th root leads to a simple result that aids in determining convergence conditions.

p-Series Test

The p-Series Test is crucial for examining convergence based on the terms' powers. It is used for series in the form of \( \sum_{k=1}^{\infty} \frac{1}{k^{p}} \), and it states:

• A series converges if \( p > 1 \).
• A series diverges if \( p \leq 1 \).

In this problem, for the boundary \( |z| = 1 \), our series becomes \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \) on evaluating the absolute values. Here, \( p = 2 \), which is greater than 1, confirming that this series converges.

This application of the p-Series Test helps conclude that: for any \( z \) on the unit circle, the power series \( \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} \) converges.

Complex Analysis

Complex analysis provides the foundation for understanding convergence in the complex plane. With power series involving a complex variable \( z \), we use methods from complex analysis to determine the behavior of the series.

The concept of convergence, whether inside or on the boundary of the circle of convergence, is crucial in complex variables. For a series \( \sum_{k=1}^{\infty} a_k z^k \), the point \( |z| < R \) shows consistent behavior straightforwardly determined by real convergence tests like the p-series or root test. However, it is at \( |z| = R \) where the depth of complex analysis is revealed as it requires tighter justification for convergence, like symmetry or periodicity of the unit circle often examined with concepts like residues or Laurent series.

In essence, complex analysis equips us with the means to handle and solve power series convergence problems in a much more comprehensive way.