Answer
Divergent
Work Step by Step
Alternating series test:
Suppose that we have series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$.
Then if the following two condition are satisfied the series is convergent.
1. $\lim\limits_{n \to \infty}b_{n}=0$
2. $b_{n}$ is a decreasing sequence.
In the given problem, $b_{n}=\frac{n^{n}}{n!}$
$\frac{n^{n}}{n!}=\frac{n.n.n.......n}{1.2.3.....n}$
Clearly, the numerator is larger than denominator because all the factors in the numerator are $n$ and there is only one factor of $n$ in the denominator.
Thus, the limit does not exist.
Hence, the given series is divergent by divergence test.
