Problem 29 - A First Course in Complex Analysis with Applications

Question

Find the circle and radius of convergence of the given power series. $$ \sum_{k=0}^{\infty} \frac{(2 k) !}{(k+2)(k !)^{2}}(z-i)^{2 k} $$

Step-by-Step Solution

Answer

The circle is centered at \(z = i\) with an infinite radius of convergence.

Step 1: Identify the General Term

The given power series is \( \sum_{k=0}^{\infty} \frac{(2 k)!}{(k+2)(k!)^{2}} (z-i)^{2k} \). Each term in the series can be written as \( a_k (z-i)^{2k} \), where \( a_k = \frac{(2k)!}{(k+2)(k!)^{2}} \).

Step 2: Use the Ratio Test for Convergence

The Ratio Test checks for convergence by evaluating \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Here, \( a_{k+1} = \frac{(2(k+1))!}{(k+3)((k+1)!)^{2}} \). Calculate \( \frac{a_{k+1}}{a_k} = \frac{(2(k+1))(2k+1)}{(k+3)} \left( \frac{k!}{(k+1)!} \right)^{2} \).

Step 3: Simplify the Ratio Expression

Simplify the expression \( \frac{(2k+2)(2k+1)}{(k+3)(k+1)^2} \). This simplifies to \( \frac{(2k+2)(2k+1)}{(k+3)(k+1)^2} \approx \frac{4}{k} \) as \( k \to \infty \). The limit of this expression as \( k \to \infty \) approaches zero.

Step 4: Determine Convergence Radius

Using the Ratio Test result, since \( \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = 0 \), the radius of convergence \( R \) is \( \infty \). Hence the power series converges for all \( z \).

Step 5: Identify the Circle

The series is of the form \((z-i)^{2k}\), centered at \(z = i\). The circle of convergence is centered at \(z = i\) with an infinite radius, meaning it converges everywhere in the complex plane.

Key Concepts

Circle of Convergence

When dealing with a power series, finding the circle of convergence is essential to understanding the series' behavior. The circle of convergence refers to a region in the complex plane within which a power series converges.
A power series in the form \( \sum_{k=0}^{\infty} a_k (z - c)^k \) is centered at \( z = c \). Thus, it converges inside a circle (or disk in some cases) around this center.
For the series \( \sum_{k=0}^{\infty} \frac{(2 k)!}{(k+2)(k!)^{2}}(z-i)^{2 k} \), the circle of convergence is centered at \( z = i \). Additionally, if the radius of convergence is infinite, the circle of convergence extends everywhere across the complex plane, implying convergence for all complex numbers \( z \).
In practical terms, it means:

• The series converges for all values of \( z \), not just around its center.
• When visualized, the circle shows where the series behaves consistently, providing insight into function analysis in complex planes.

Radius of Convergence

The radius of convergence \( R \) is a crucial factor in determining where a power series converges. It signifies the distance from the center \( z = c \) within which the series converges.
To find the radius, mathematicians often deploy tests such as the Ratio Test or Root Test, which involve examining the series' coefficients.
For our series, the Ratio Test showed that

• The expression \( \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = 0 \)
• This result indicates that the series converges for any value of \( z \), leading to a radius of convergence \( R = \infty \).

This infinite radius implies that the series has absolute convergence throughout the entire complex plane, rather than just in a finite bound around the center. Such clear, global convergence is a unique and helpful trait of some power series, indicating their stability and applicability across various mathematical sectors.

Ratio Test

The Ratio Test is a staple technique used to assess the convergence of infinite series. This test involves comparing the ratio of successive terms in a series.
The main idea is to evaluate:

• \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \)
• If this limit is less than 1, the series converges.
• If greater than 1, the series diverges.
• If equal to 1, the test is inconclusive.

For the given problem, we determined that:
By calculating the terms related specifically to the series' formula, the ratio showed:

• Convergence throughout all values, as \( \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = 0 \).
• Thus confirming the broad convergence found through the infinite radius of convergence.

The simplicity and widespread applicability of the Ratio Test make it a preferred choice, ensuring students can confidently determine whether a series behaves predictably within the given boundaries.