Could someone please verify my solution to the following question?
Question: Does $\sum_{n \in \mathbb{N}} (-1)^n \dfrac{n!}{n^n}$ converge?
Answer: I have verified with WolframAlpha that it does converge. I wish to know if my proof is correct:
Define $a_n = \dfrac{(-1)^n n!}{n^n}$. Then $$\left|\dfrac{a_{n+1}}{a_n}\right|= \dfrac{(n+1)!n^n}{(n+1)^{n+1} n!} = (1+1/n)^{-n}.$$
Hence, we see that $\lim\limits_{n\rightarrow\infty} \left|\dfrac{a_{n+1}}{a_n}\right|= \dfrac{1}{e}<1$ from which we deduce that the series is absolutely convergent by the ratio test. Hence the series in question converges.
Your proof is fine. You may also notice that for any $n\geq 1$ $$ \frac{n!}{n^n} = \int_{0}^{+\infty} n z^n e^{-nz}\,dz $$ so $$ \sum_{n\geq 0}(-1)^n\frac{n!}{n^n} x^n = 1-x\int_{0}^{+\infty}\frac{z e^z}{(e^z+xz)^2}\,dz$$ where the radius of convergence of the LHS is given by the supremum of $\rho\in\mathbb{R}^+$ such that $e^z > \rho z$ for any $z>0$, i.e. by $e$ as also shown by Stirling's approximation.