Four series involving logarithms

(1) $\sum\frac{1}{(\ln n)^n}$. By Cauchy's root test:
$$\sqrt[n]{\frac{1}{(\ln n)^n}}=\frac{1}{\ln n}\longrightarrow0<1.$$
The series converges.

(2) $\sum\frac{H_n}{\ln(n!)}$. First compare $\ln(n!)$ with $n\ln n$: for $1\le k\le n$, $\ln k\le\ln n$, hence
$$\ln(n!)=\sum_{k=1}^{n}\ln k\le n\ln n.$$
Therefore
$$\frac{H_n}{\ln(n!)}\ge\frac{H_n}{n\ln n}.$$
Moreover $H_n\ge\ln n$ (comparison of the harmonic sum with $\int_1^n\frac{dx}{x}$), so
$$\frac{H_n}{n\ln n}\ge\frac{\ln n}{n\ln n}=\frac1n.$$
Since $\sum\frac1n$ diverges, by comparison the series diverges.

(3) $\sum\frac1n\ln(n)\ln(1+1/n)$. Since $\ln(1+1/n)\sim\frac1n$ as $n\to\infty$, the general term satisfies
$$u_n=\frac{\ln n}{n}\cdot\ln\!\left(1+\frac1n\right)\sim\frac{\ln n}{n^2}.$$
Now $\frac{\ln n}{n^2}=o\!\left(\frac{1}{n^{3/2}}\right)$ (since $\ln n/\sqrt n\to0$), and $\sum n^{-3/2}$ converges (Riemann, $\alpha=3/2>1$). By comparison, the series converges.

(4) $\sum\frac{1}{(\ln n)^{\ln n}}$. Rewrite
$$\frac{1}{(\ln n)^{\ln n}}=\frac{1}{\exp(\ln n\cdot\ln\ln n)}=\frac{1}{n^{\ln\ln n}}.$$
Since $x\mapsto\ln\ln x$ is increasing and tends to $+\infty$ on $[e,+\infty[$, there exists $N$ with $\ln\ln n\ge2$ for all $n\ge N$. For such $n$,
$$u_n\le\frac{1}{n^2},$$
the general term of a convergent Riemann series. By comparison, the series converges.