We are tasked with finding the center and radius of convergence of the series:

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

Step 1: Express the series in the general form

The general form of a power series is:

$\sum_{n=0}^{\infty} a_n z^n$

In our case, the series is:

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

Notice that the powers of $z$ are $3n$, not just $n$, so the series is not in the standard form of a power series. To proceed, let’s make a substitution to express the series in a way that allows us to use standard tests for convergence.

Step 2: Define a new variable $w = z^3$

Let $w = z^3$, so that the series becomes:

$\sum_{n=1}^{\infty} \frac{3n(3n - 1)}{n^n} w^n$

Now, the series is in the standard form:

$\sum_{n=1}^{\infty} a_n w^n$

where:

$a_n = \frac{3n(3n - 1)}{n^n}$

Step 3: Apply the root test to determine the radius of convergence

The root test tells us that the radius of convergence $R_w$ of the series $\sum a_n w^n$ is given by:

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

Thus, we need to find:

$\limsup_{n \to \infty} \sqrt[n]{\left| \frac{3n(3n - 1)}{n^n} \right|}$

Step 4: Simplify the expression for $a_n$

First, simplify the general term:

$|a_n| = \frac{3n(3n - 1)}{n^n}$

As $n$ grows large, $3n(3n - 1)$ behaves asymptotically as $9n^2$. Therefore:

$|a_n| \sim \frac{9n^2}{n^n}$

Now, we need to evaluate the limit of $\sqrt[n]{|a_n|}$:

[ \lim_{n \to \infty} \sqrt[n]{\frac{9n^2}{n^n}} = \lim_{n \to \infty} \frac{\sqrt[n]{9n^2}}{n} = \lim_{n \to \infty} \frac{9^{1/n} n^{2/n}}{n} ]

As $n \to \infty$, $9^{1/n} \to 1$ and $n^{2/n} \to 1$, so:

$\lim_{n \to \infty} \frac{9^{1/n} n^{2/n}}{n} = 0$

Thus, the radius of convergence in terms of $w$ is infinite:

$R_w = \infty$

Step 5: Convert back to $z$

Since $w = z^3$, the radius of convergence in terms of $z$ is:

[ R_z = \sqrt[3]{R_w} = \sqrt[3]{\infty} = \infty ]

Conclusion:

• The center of convergence is $z = 0$ (since the series is centered around $w = 0$).
• The radius of convergence is $\infty$.

If you need further details or have any questions, feel free to ask!

Related questions:

1. How does the root test apply to different power series?
2. What is the significance of the radius of convergence in complex analysis?
3. How can we find the radius of convergence of a series that involves both powers of $z$ and other functions?
4. What happens if we apply the ratio test instead of the root test for this series?
5. How does the growth of $n^n$ affect the convergence of series like this?

Tip: When dealing with series where the powers of $z$ are non-linear (like $z^{3n}$ instead of $z^n$), always make a substitution to simplify the series into a more familiar form.