How Do You Find the Disk of Convergence?

The disk of convergence for a power series is the region in the complex plane where the series converges. To find it, you need to identify two key components: its center and its radius of convergence.

Understanding the Disk of Convergence

A power series is generally expressed in the form:

$$
\sum_{n=0}^{\infty} c_n (z-a)^n = c_0 + c_1(z-a) + c_2(z-a)^2 + \ldots
$$

Here:

• $c_n$ are the coefficients of the series.
• $z$ is the complex variable.
• $a$ is the center of the power series (and thus the center of the disk of convergence).

The series will converge for all values of $z$ that lie inside an open disk centered at $a$. This region is called the disk of convergence. If the power series is expanded around the point a and the radius of convergence is r, then the set of all points z such that |z − a| = r is a circle called the boundary of the disk of convergence.

Steps to Find the Disk of Convergence

Finding the disk of convergence primarily involves determining its radius.

Step 1: Identify the Center of the Power Series (a)

The center of the power series is the point around which the series is expanded. In the general form $\sum_{n=0}^{\infty} c_n (z-a)^n$, the center is simply a.

• If the series is $\sum_{n=0}^{\infty} c_n z^n$, the center is $a=0$.
• If the series is $\sum_{n=0}^{\infty} c_n (z - (2+3i))^n$, the center is $a=2+3i$.

Step 2: Determine the Radius of Convergence (r)

The radius of convergence, denoted by r (or R), defines the size of the disk. It can be found using the Ratio Test or the Root Test.

A. Using the Ratio Test

The Ratio Test is the most commonly used method for finding the radius of convergence. For a series $\sum_{n=0}^{\infty} c_n (z-a)^n$, the radius of convergence r is given by:

$$
r = \lim_{n \to \infty} \left| \frac{cn}{c{n+1}} \right|
$$

provided this limit exists. If the limit is 0, the radius of convergence is $\infty$ (series converges everywhere). If the limit is $\infty$, the radius of convergence is 0 (series converges only at its center).

Procedure for the Ratio Test:

1. Identify the coefficients $c_n$.
2. Form the ratio $\left| \frac{cn}{c{n+1}} \right|$.
3. Calculate the limit of this ratio as $n \to \infty$. This limit is r.

B. Using the Root Test

The Root Test is another powerful method, particularly useful when $cn$ involves powers of n. For a series $\sum{n=0}^{\infty} c_n (z-a)^n$, the radius of convergence r is given by:

$$
r = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|c_n|}}
$$

provided the limit exists. Similar to the Ratio Test, if $\lim_{n \to \infty} \sqrt[n]{|cn|} = 0$, then $r = \infty$. If $\lim{n \to \infty} \sqrt[n]{|c_n|} = \infty$, then $r = 0$.

Procedure for the Root Test:

1. Identify the coefficients $c_n$.
2. Calculate $\sqrt[n]{|c_n|}$.
3. Calculate the limit of this expression as $n \to \infty$. Let this limit be L.
4. The radius of convergence is $r = 1/L$.

Choosing Between Ratio and Root Test:
| Feature | Ratio Test | Root Test |
| :---------------- | :---------------------------------------------- | :-------------------------------------------- |
| Formula | $\lim_{n \to \infty} \left| \frac{cn}{c{n+1}} \right|$ | $1 / \lim_{n \to \infty} \sqrt[n]{|c_n|}$ |
| When to Use | Most common, especially with factorials ($n!$) or products. | When $c_n$ contains terms raised to the power of n (e.g., $n^n$, $(2n)^n$). |
| Ease of Use | Often simpler algebraically. | Can be more complex if $\sqrt[n]{|c_n|}$ is hard to evaluate. |

Step 3: Define the Disk of Convergence

Once you have the center a and the radius of convergence r, the disk of convergence is the set of all complex numbers $z$ such that |z - a| < r. This represents an open disk in the complex plane.

• If $r = \infty$: The series converges for all $z \in \mathbb{C}$. The disk of convergence is the entire complex plane.
• If $r = 0$: The series converges only at its center, $z=a$. The disk of convergence is just a single point.

Practical Example: Finding the Disk of Convergence

Let's find the disk of convergence for the power series:

$$
\sum_{n=1}^{\infty} \frac{(z - 3i)^n}{n \cdot 2^n}
$$

Step 1: Identify the Center (a)

Comparing with the general form $\sum_{n=0}^{\infty} c_n (z-a)^n$, we see that the series is centered at $a = 3i$.

Step 2: Determine the Radius of Convergence (r)

Here, $c_n = \frac{1}{n \cdot 2^n}$. We will use the Ratio Test.

1. Identify $cn$ and $c{n+1}$:
   $cn = \frac{1}{n \cdot 2^n}$
   $c{n+1} = \frac{1}{(n+1) \cdot 2^{n+1}}$

2. Form the ratio $\left| \frac{cn}{c{n+1}} \right|$:
   $$
   \begin{align}
   \left| \frac{cn}{c{n+1}} \right| &= \left| \frac{\frac{1}{n \cdot 2^n}}{\frac{1}{(n+1) \cdot 2^{n+1}}} \right| \
   &= \left| \frac{(n+1) \cdot 2^{n+1}}{n \cdot 2^n} \right| \
   &= \left| \frac{(n+1) \cdot 2^n \cdot 2}{n \cdot 2^n} \right| \
   &= \left| \frac{2(n+1)}{n} \right| \
   &= 2 \left| \frac{n+1}{n} \right| \
   &= 2 \left| 1 + \frac{1}{n} \right|
   \end{align}
   $$

3. Calculate the limit as $n \to \infty$:
   $$
   r = \lim_{n \to \infty} 2 \left| 1 + \frac{1}{n} \right| = 2 (1 + 0) = 2
   $$
   So, the radius of convergence is $r = 2$.

Step 3: Define the Disk of Convergence

With the center $a = 3i$ and the radius $r = 2$, the disk of convergence is the set of all complex numbers $z$ such that:

$$
|z - 3i| < 2
$$

This represents an open disk centered at $3i$ with a radius of $2$.