Cauchy ratio test yields 1 (so it's inconclusive). I have tried this:
$$\frac{1}{n \log^2n}=\frac{1}{n \log n \log n}=\frac{1}{\log n^n \log n}\geq \frac{1}{\log n^n -n} \approx \frac{1}{\log n!} $$
Now, since $\sum 1/\log n!$ diverges, the original series must diverge too. But Wolfram Alpha says it's convergent. How did I go wrong and how can I solve this?
You might find it easier to apply the Cauchy condensation test:
$$\sum_{n=2}^\infty\frac1{n\log^2n}\le\sum_{n=1}^\infty\frac{2^n}{2^n\log^22^n}=\sum_{n=1}^\infty\frac1{n^2\log^22}$$
Fundamentally, what went wrong was your inequality chain. We have that $$\frac{1}{\ln \left(x^x\right)-x} \gg \frac{1}{\ln \left(x^x\right)\ln x}$$ contrary to what your inequality chain would imply.
