How do I prove the divergence of this series? How do I prove that $\displaystyle\sum_{n\geq 1}\frac {1}{\ln^2n}$ is a divergent series?

Short answer: Since $n > \log n$ for all $n \in \mathbb R^+$, we know that $\frac{1}{n (\log n)} \leq \frac{1}{(\log n)^2}$ for all $n \in \mathbb R^+$.
We know that $\sum_{n=1}^\infty \frac{1}{n (\log n)}$ diverges, so by the comparison test, so does $\sum_{n=1}^\infty \frac{1}{(\log n)^2}$.
Longer answer: Here's some information that explains the motivation of comparing the series with $\sum_{n=1}^\infty \frac{1}{n (\log n)}$. It is a commonly known fact that all the following series diverge:


*

*$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ 

*$\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n)}$ 

*$\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)}$

*$\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)(\log \log \log n)}$

*and so on


You can prove those by using the integral test (or the Cauchy condensation test, if you prefer). When you apply either of these tests, each series reduces to the series above it. We know the first one (which is the harmonic series) diverges, so by induction, all the other series diverge as well.
