Let's go through the solution step-by-step again to identify the error and correct it.

The series given is: $\sum_{k=1}^\infty \frac{3n - 2}{5^n}$

First, let's correct the indices and properly handle each part of the series separately.

Step 1: Split the Series

$\sum_{n=1}^\infty \frac{3n - 2}{5^n} = \sum_{n=1}^\infty \frac{3n}{5^n} - \sum_{n=1}^\infty \frac{2}{5^n}$

Step 2: Evaluate Each Series Separately

Sum 1: $\sum_{n=1}^\infty \frac{2}{5^n}$

This is a geometric series with the first term $a = \frac{2}{5}$ and common ratio $r = \frac{1}{5}$: $\sum_{n=1}^\infty \frac{2}{5^n} = \frac{\frac{2}{5}}{1 - \frac{1}{5}} = \frac{\frac{2}{5}}{\frac{4}{5}} = \frac{2}{4} = \frac{1}{2}$

Sum 2: $\sum_{n=1}^\infty \frac{3n}{5^n}$

We use the formula for the sum of a series involving $n$ multiplied by a geometric progression: $\sum_{n=1}^\infty \frac{n}{5^n} = \frac{r}{(1 - r)^2} \quad \text{where} \quad r = \frac{1}{5}$

Thus: $\sum_{n=1}^\infty \frac{n}{5^n} = \frac{\frac{1}{5}}{(1 - \frac{1}{5})^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{\frac{1}{5}}{\frac{16}{25}} = \frac{25}{80} = \frac{5}{16}$

Now, multiply this result by 3: $\sum_{n=1}^\infty \frac{3n}{5^n} = 3 \times \frac{5}{16} = \frac{15}{16}$

Step 3: Combine the Results

$\sum_{n=1}^\infty \frac{3n - 2}{5^n} = \frac{15}{16} - \frac{1}{2} = \frac{15}{16} - \frac{8}{16} = \frac{7}{16}$

Thus, the correct sum of the series is: $\boxed{\frac{7}{16}}$

Do you have any questions or would you like more details on any step?

Here are 5 more questions you can ask:

1. How do you find the sum of a geometric series?
2. Can you explain the steps to derive the sum formula for $\sum_{n=1}^\infty \frac{n}{5^n}$?
3. What are other methods to solve series involving factorials?
4. How do you approach solving an alternating series?
5. Can you explain the concept of convergence in series?

Tip: Carefully checking each step of a series evaluation can help catch and correct errors in the calculation.