The given series is:

$\sum_{n=0}^{\infty} \frac{(z - 2i)^n}{n^n}$

To find the radius of convergence of this power series, we can apply the Root Test (Cauchy-Hadamard theorem). The radius of convergence $R$ is given by:

$\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$

where $a_n = \frac{(z - 2i)^n}{n^n}$.

Step 1: Find the general term

The general term is:

$a_n = \frac{(z - 2i)^n}{n^n}$

We can express this as:

$a_n = \left( \frac{z - 2i}{n} \right)^n$

Step 2: Apply the Root Test

We now apply the root test to this series. Taking the $n$-th root of $a_n$:

[ \sqrt[n]{|a_n|} = \sqrt[n]{\left| \frac{(z - 2i)^n}{n^n} \right|} ]

Simplifying:

$\sqrt[n]{|a_n|} = \left| \frac{z - 2i}{n} \right|$

This simplifies further to:

$\sqrt[n]{|a_n|} = \frac{|z - 2i|}{n}$

Step 3: Find the limit

Taking the limit as $n \to \infty$:

$\lim_{n \to \infty} \frac{|z - 2i|}{n} = 0$

Since this limit is 0, the series converges for all $z$. Therefore, the radius of convergence $R$ is infinite.

Final Answer:

The radius of convergence is $\boxed{R = \infty}$.

Would you like more details on how the Root Test works, or any additional explanations?

Here are 5 related questions you could explore:

1. How does the Root Test compare to the Ratio Test for finding the radius of convergence?
2. What is the significance of an infinite radius of convergence for a power series?
3. How would the radius of convergence change if the general term were $\frac{(z - 2i)^n}{n!}$?
4. What are some common series with finite radii of convergence?
5. How do singularities of a function affect the radius of convergence of its series expansion?

Tip: When applying the Root Test, the behavior of $a_n$ as $n \to \infty$ is key to determining convergence.