How to prove the convergence of the series: $$\sum_n \dfrac{(\ln(n))^n}{n!}$$
I tried to use the d'Alembert's rule: If we write $u_n = \dfrac{(\ln(n))^n}{n!}$ then we get
$$\dfrac{u_{n+1}}{u_n} = \dfrac{1}{n+1} \times \dfrac{(\ln(n+1))^{n+1}}{(\ln(n))^n}$$
Is that possible to show that $\dfrac{u_{n+1}}{u_n} \rightarrow l <1$ ?
Since $\log(n)\leq \sqrt{n}$ and $n!\geq \frac{n^n}{e^n}$ we have $$ 0\leq \sum_{n\geq 1}\frac{\log(n)^n}{n!}\leq \sum_{n\geq 1}\frac{e^n}{n^{n/2}}=\sum_{n\geq 1}\frac{1}{\left(\frac{1}{e}\sqrt{n}\right)^n} $$ and convergence by asymptotic comparison with a geometric series.
Computations show $$\begin{aligned} \ln(\dfrac{u_{n+1}}{u_n}) = (n+1)\ln\ln(n+1)-n\ln\ln n - \ln (n+1)\sim-\ln n \end{aligned}$$
so $\dfrac{u_{n+1}}{u_n} \to 0$. This yields convergence of the series.
Another way of getting convergence could resort to Stirling's estimate: $$\dfrac{(\ln(n))^n}{n!}\sim \left(\frac{e\ln n}{n} \right)^n \frac{1}{\sqrt{2\pi n}}$$
For large $n$, $\displaystyle \frac{\ln n}n\leq \frac{1}{e^2} $, hence $\displaystyle \frac{e\ln n}{n}\leq \frac{1}{e}$.
Thus $$\dfrac{(\ln(n))^n}{n!} = O\left(\frac{1}{e^n} \right)$$ The series converges.
From $$\frac{u_{n+1}}{u_n}=\frac{1}{n+1}\frac{(\ln{(n+1)})^{n+1}}{(\ln{n})^n}=\frac{\ln{(n+1)}}{n+1}\left(\frac{\ln{(n+1)}}{\ln{n}}\right)^n=\\ \frac{\ln{(n+1)}}{n+1}\left(1+\frac{\ln{\left(1+\frac{1}{n}\right)}}{\ln{n}}\right)^n=...$$ Note $a_n=\frac{\ln{\left(1+\frac{1}{n}\right)}}{\ln{n}} \rightarrow 0, n\rightarrow\infty$ $$...=\frac{\ln{(n+1)}}{n+1}\left(\left(1+a_n\right)^{\frac{1}{a_n}}\right)^{a_nn} \rightarrow 0, n\rightarrow\infty$$ because $$\lim\limits_{n\rightarrow\infty}\frac{\ln{(n+1)}}{n+1}=0$$ $$\lim\limits_{n\rightarrow\infty}\left(1+a_n\right)^{\frac{1}{a_n}}=e \Rightarrow e-\varepsilon < \left(1+a_n\right)^{\frac{1}{a_n}} < e+\varepsilon \tag{1}$$ $$\lim\limits_{n\rightarrow\infty}a_nn=\lim\limits_{n\rightarrow\infty}n\frac{\ln{\left(1+\frac{1}{n}\right)}}{\ln{n}}=\lim\limits_{n\rightarrow\infty}\frac{\ln{\left(1+\frac{1}{n}\right)^n}}{\ln{n}}=\frac{1}{\lim\limits_{n\rightarrow\infty} \ln{n}}=0 \tag{2}$$ from $(1)$ and $(2)$ $$\left(e-\varepsilon\right)^{a_nn} < \left(\left(1+a_n\right)^{\frac{1}{a_n}}\right)^{a_nn} < \left(e+\varepsilon\right)^{a_nn} \Rightarrow \lim\limits_{n\rightarrow\infty}\left(\left(1+a_n\right)^{\frac{1}{a_n}}\right)^{a_nn}=1$$
Write
$$\frac{u_{n+1}}{u_n} = \frac{\ln(n+1)}{n+1}\times \left(\frac{\ln(n+1)}{\ln n}\right)^n$$
Since $\ln(n+1) = 2\ln \sqrt{n+1} < 2\sqrt{n+1}$, it follows from the squeeze theorem that
$$\lim_{n\to \infty} \frac{\ln(n+1)}{n+1} = 0$$
Now
$$\frac{\ln(n+1)}{\ln n} = 1 + \frac{\ln(n+1) - \ln n}{\ln n} = 1 + \frac{\ln(1 + \frac{1}{n})}{\ln n}$$
so by the binomial theorem,
$$\left(\frac{\ln(n+1)}{\ln n}\right)^n > 1 + n\frac{\ln(1 + \frac{1}{n})}{\ln n} = 1 + \frac{\ln(1 + \frac{1}{n})}{\frac{1}{n}}\frac{1}{\ln n}$$
The limit of $n\dfrac{\ln(1 + \frac{1}{n})}{\ln n}$ is $0$ since $\lim\limits_{n\to \infty} \dfrac{1}{\ln n} = 0$ and $\dfrac{\ln(1 + \frac{1}{n})}{\frac{1}{n}}$ is bounded by $1$. As
$$1 + n\frac{\ln(1 + \frac{1}{n})}{\ln n} < \left(\frac{\ln(n+1)}{\ln n}\right)^n < 1$$
the squeeze theorem yields $\lim\limits_{n\to \infty} \left(\dfrac{\ln(n+1)}{\ln n}\right)^n = 1$. Consequently, $\lim\limits_{n\to \infty} \dfrac{u_{n+1}}{u_n} = 0$.
