convergence divergence - Is $sum_{n=2}^{infty} frac{1}{nlog^2 {n}}$ convergent or divergent?

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?

Answer

You might find it easier to apply the Cauchy condensation test:

$$\sum_{n=2}^\infty\frac1{n\log^2n}\le\sum_{n=1}^\infty\frac{2^n}{2^n\log^22^n}=\sum_{n=1}^\infty\frac1{n^2\log^22}$$