Another simple series convergence question: $\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}$

If $a_n$ is eventually nonincreasing, then $\sum a_n$ converges iff $\sum 2^na_{2^n}$ converges (Cauchy condensation test). This takes us from $$\sum\frac 1{n\ln n \ln\ln n}$$ to $$\sum\frac {2^n}{2^n\cdot n\ln 2 \cdot (\ln n + \ln\ln 2)}=\sum\frac {1}{n\ln 2 \cdot (\ln n + \ln\ln 2)}$$ and then to $$\sum\frac {2^n}{2^n\ln 2 \cdot (n \ln 2 + \ln\ln 2)}=\sum\frac {1}{n\ln^2 2 + \ln2\cdot\ln\ln 2}$$ which is essentially the harmonic series.

Apply directly integral test

$$\int_3^{\infty}\frac{(\ln(\ln x))'}{ \ln(\ln x)}\mathrm{dx}= \left[\ln(\ln(\ln(x)))\right]_3^{\infty}\longrightarrow \infty$$