The given series is:

$\sum_{n=1}^{\infty} \frac{9 \cdot 2^{3n-3} - 4 \cdot 5^{n+2}}{3^{2n}}$

We will break down this problem step by step to determine whether the series converges or diverges, and if it converges, we'll find its sum.

Step 1: Rewrite the general term

First, rewrite the expression inside the summation to make it more manageable. Let's split the term in two parts:

$\frac{9 \cdot 2^{3n-3} - 4 \cdot 5^{n+2}}{3^{2n}} = \frac{9 \cdot 2^{3n-3}}{3^{2n}} - \frac{4 \cdot 5^{n+2}}{3^{2n}}$

Simplifying the first part:

$\frac{9 \cdot 2^{3n-3}}{3^{2n}} = 9 \cdot \frac{2^{3n-3}}{3^{2n}}$

We can rewrite the exponents of 2 and 3:

$= 9 \cdot \frac{2^{3n}}{2^3} \cdot \frac{1}{3^{2n}} = 9 \cdot \frac{2^{3n}}{8 \cdot 3^{2n}} = \frac{9}{8} \cdot \left(\frac{2^3}{3^2}\right)^n = \frac{9}{8} \cdot \left(\frac{8}{9}\right)^n$

So the first term becomes:

$\frac{9}{8} \cdot \left(\frac{8}{9}\right)^n$

Simplifying the second part:

$\frac{4 \cdot 5^{n+2}}{3^{2n}} = 4 \cdot \frac{5^{n+2}}{3^{2n}} = 4 \cdot \frac{5^n \cdot 25}{3^{2n}} = 100 \cdot \left(\frac{5}{9}\right)^n$

Thus, the original series becomes:

$\sum_{n=1}^{\infty} \left( \frac{9}{8} \cdot \left( \frac{8}{9} \right)^n - 100 \cdot \left( \frac{5}{9} \right)^n \right)$

Step 2: Check for convergence

This is now a sum of two geometric series. The general form of a geometric series is:

$\sum_{n=1}^{\infty} ar^n = \frac{ar}{1-r} \quad \text{if} \ |r| < 1$

First geometric series:

The first series is:

$\frac{9}{8} \sum_{n=1}^{\infty} \left( \frac{8}{9} \right)^n$

Here, $a = \frac{9}{8}$ and $r = \frac{8}{9}$. Since $\left| \frac{8}{9} \right| < 1$, the series converges, and its sum is:

$S_1 = \frac{\frac{9}{8} \cdot \frac{8}{9}}{1 - \frac{8}{9}} = \frac{\frac{9}{8} \cdot \frac{8}{9}}{\frac{1}{9}} = 1 \cdot 9 = 9$

Second geometric series:

The second series is:

$100 \sum_{n=1}^{\infty} \left( \frac{5}{9} \right)^n$

Here, $a = 100$ and $r = \frac{5}{9}$. Since $\left| \frac{5}{9} \right| < 1$, the series converges, and its sum is:

$S_2 = \frac{100 \cdot \frac{5}{9}}{1 - \frac{5}{9}} = \frac{100 \cdot \frac{5}{9}}{\frac{4}{9}} = \frac{500}{4} = 125$

Step 3: Find the total sum

The total sum of the series is:

$S = 9 - 125 = -116$

Conclusion:

The series converges, and its sum is -116.

Do you have any questions, or would you like further details on any part of this solution?

Here are 5 related questions to expand your understanding:

1. How do you determine if a geometric series converges or diverges?
2. What is the general formula for the sum of a geometric series?
3. Can you explain why the ratio test is useful for series like this?
4. What would happen if the common ratio $r$ in either series was greater than 1?
5. How do we handle series that do not resemble geometric series?

Tip: When dealing with series, always check if they resemble geometric or known convergent forms before applying more complex tests.