Convergence of the function series $\sum \frac{n!}{(nx)^n}$ for $x<0$ We want to determine the $x<0$ in $\mathbb{R}$ such that
$$\sum _{n=1}^{\infty} \dfrac{n!}{(nx)^n}$$
converges. The textbook says that the solution is $x<-1/e$. What I thought is that, since $x<0$, we have that $x^n>0$ if $n$ is even and $x^n<0$ if $n$ is odd, so we can rewrite the series as
$$\sum _{n=1}^{\infty} (-1)^n \dfrac{n!}{(nx)^n} \qquad \qquad \text{for } x>0$$
And because of the Leibniz test, the series converges if the sequence $(nx)^{-n}n!$ is decreasing and infinitesimal, but we have that
$$\lim _{n \rightarrow +\infty} \dfrac{n!}{(nx)^n}=\lim _{n \rightarrow +\infty} \dfrac{n!}{n^n} \cdot \dfrac{1}{x^n}=0 \Longleftrightarrow \lim _{n\rightarrow +\infty} \dfrac{1}{x^n} < +\infty \Longrightarrow x \ge1$$
Because $n!/n^n \rightarrow 0$ for $n \rightarrow +\infty$, but this seems to go nowhere. What can I do?

Because $x \in \mathbf{R}$, a good idea is to study the absolute convergence of the series. Let:
$$a_n=\left|\frac{n!}{(x\cdot n)^n}\right|$$
Then, we can applay the root-criteria. In particluar, we have to determine the following limit:
$$\lim_{n\to +\infty}\sqrt[n]{a_n}=\lim_{n\to +\infty}\sqrt[n]{\left|\frac{n!}{(x\cdot n)^n}\right|}=\lim_{n\to +\infty}\frac{\sqrt[n]{\frac{n!}{n^n}}}{|x|}=(*)$$
Theorem: Let $\{b_n\}_{n\geq n_0}$ a real sequence, such that $a_n>0$ at least definitely. If  $\lim_{n\to +\infty}\sqrt[n]{b_n}$ and $\lim_{n\to +\infty}\frac{b_{n+1}}{b_n}$ exist, then their limits are equal. Here a post where I used this tool.
Let $b_n=\sqrt[n]{\frac{n!}{n^n}}$, we have:
$$\lim_{n\to +\infty}\frac{b_{n+1}}{b_n}=\lim_{n \to +\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to +\infty}\frac{(n+1)!}{(n+1)\cdot n!}\cdot\left(\frac{n}{n+1}\right)^n=\lim_{n\to +\infty}\left(1-\frac{1}{n}\right)^n=e^{-1}=\frac{1}{e}$$
So, the first limits becomes:
$$(*)=\lim_{n\to +\infty}\frac{\sqrt[n]{\frac{n!}{n^n}}}{|x|}=\frac{1}{e\cdot|x|}$$
By root-criteria, the series converges if and only if:
$$\frac{1}{e\cdot|x|}<1\implies \frac{1}{e}<|x|\implies x\in \left(-\infty, -\frac{1}{e}\right)$$
