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 $a_{n+1}$: = \frac{(-1)^{n+2} (n+1)^{2n+2}}{(n+1)!}$$
2. Form the ratio $\frac{a_{n+1}}{a_n}$: = \frac{(n+1)^{2n+2}}{(n+1) n^{2n}}$$

After simplifying and taking the limit,