Answer
Converges
Work Step by Step
Let us consider an alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ that can be convergent , when it satisfy the following two conditions such as: 1) The magnitude of terms must form a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=0$
Here, in this problem we have $a_k=\dfrac{k !}{k^k}$
a) In the given sequence, $a_k=\dfrac{k !}{k^k}$ , and $a_{k+1}=\dfrac{(k+1) !}{(k+1)^{k+1}}$ and $\dfrac{(k+1) !}{(k+1)^{k+1}} \leq \dfrac{k !}{k^k}$
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{k !}{k^k}=0$
This implies that the given series converges.
