Convergence of the serie $\sum \dfrac{(\ln(n))^n}{n!}$ 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 $\dfrac{u_{n+1}}{u_n} \rightarrow l <1$ ?


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, $\lim\limits_{n\to \infty} \dfrac{u_{n+1}}{u_n} = 0$.
