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To determine whether the series 

\[
\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{2n}}{n!}
\]

is absolutely convergent or divergent, we can use the **Ratio Test**.

The Ratio Test states that for a series \(\sum a_n\), we calculate the limit 

\[
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]

- If \(L < 1\), the series is absolutely convergent.
- If \(L > 1\), the series is divergent.
- If \(L = 1\), the test is inconclusive.

Here, we have \(a_n = \frac{(-1)^{n+1} n^{2n}}{n!}\). Let's apply the Ratio Test to this series.

### Step 1: Compute \( \left| \frac{a_{n+1}}{a_n} \right| \)

1. **Calculate \(a_{n+1}\):**
   \[
   a_{n+1} = \frac{(-1)^{(n+1)+1} (n+1)^{2(n+1)}}{(n+1)!}
   = \frac{(-1)^{n+2} (n+1)^{2n+2}}{(n+1)!}
   \]

2. **Form the ratio \( \frac{a_{n+1}}{a_n} \):**
   \[
   \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+2} (n+1)^{2n+2}}{(n+1)!} \cdot \frac{n!}{(-1)^{n+1} n^{2n}} \right|
   = \frac{(n+1)^{2n+2}}{(n+1) n^{2n}}
   \]

After simplifying and taking the limit,

Determine whether the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{2n}}{n!} \) is absolutely convergent or divergent using the Ratio Test.

This solution analyzes whether the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{2n}}{n!} \) is absolutely convergent or divergent by applying the Ratio Test. Detailed steps are provided for simplifying terms and taking the limit, aimed at Calculus II students.

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