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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.

Determine whether the series \( \sum_{n=1}^{\infty} \frac{9 \cdot 2^{3n-3} - 4 \cdot 5^{n+2}}{3^{2n}} \) is convergent or divergent. If convergent, find the sum.

Explore the convergence of the series \( \sum_{n=1}^{\infty} \frac{9 \cdot 2^{3n-3} - 4 \cdot 5^{n+2}}{3^{2n}} \). Learn step-by-step how to determine if the series converges and find its sum if it does. This problem includes concepts of geometric series, ratio tests, and series convergence tests.

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