When does $P(x)=\sum_{n=0}^{\infty}{\frac{n!}{n^n}\cdot x^n}$ with $x\in \mathbb{C}$ converge?

Question

When does $P(x)=\sum_{n=0}^{\infty}{\frac{n!}{n^n}\cdot x^n}$ with $x\in \mathbb{C}$ converge?

My Attempt:$$\bigg |\frac{a_n}{a_{n+1}}\bigg |=\left(1+\frac{1}{n}\right)^n\to e$$ So it converges for $|x|<e$. Now I wanted to check wheter the series converges for $e$ or $-e$.

I fail to either prove the divergence or convergence. Can you help?

Answer 1

In fact $\left(1+\frac1n\right)^n<e$, and this can be used to prove by induction that $\left|e^n\frac{n!}{n^n}\right|>1$ for all $n\geq 1$. Thus when $|x|=e$ the terms don't tend to zero and it can't converge.

Answer 2

An approach without Stirling, for $x=-e$ you can just bound $\frac{n!}{n^n}$ by 1.

For $x=e$, we know that $e=\lim\limits_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n$ that sequence is increasing, so $$e^n\geq \prod\limits_{k=1}^n \left(1+\frac{1}{k}\right)^k=\frac{(n+1)^n}{n!}$$ So, we have that:

$$\frac{n! e^n}{n^n}\geq \frac{(n+1)^n}{n^n}>1$$ so series diverge