The series of 1/(ln n)^(ln n)

Rewriting. $\dfrac1{(\ln n)^{\ln n}}=\dfrac1{n^{\ln\ln n}}$.

Comparison. Since $\ln\ln n\to+\infty$, there is $n_0$ (namely $n_0=\lceil e^{e^2}\rceil$) such that $\ln\ln n\ge2$ for $n\ge n_0$. For such $n$,
$$\frac1{n^{\ln\ln n}}\le\frac1{n^2}.$$
Hence the series converges.

Remark. The exponent is not constant but eventually exceeds any fixed number, so the series is much "more summable" than any $p$-series. But note the threshold $n_0\approx1618$ already: the first terms are large, since $\ln n$ is small.