Discover whether $\sum_{n = 2}^{\infty} \frac{1}{n\log(n)}$ is convergent or not, using Cauchy I was asked  if $\sum\limits_{n=2}^{\infty} \frac{1}{n\log(n)}$ was convergent or not. I already solved this problem using the integral property, but I wanted to use Cauchy instead.
I defined $m,n \in \mathbb N$ with $m \lt n \land \exists \epsilon \gt 0$ in a way that
$$\left| \sum_{n=2}^{\infty} \frac{1}{n \log (n)} - \sum_{m=2}^{\infty} \frac{1}{m \log (m)} \right| \gt \epsilon$$
I am now stuck at this part where I would need to expend the sums and simplify them. I can’t find a way to go further from here. I don’t have the impression to actually prove anything. Am I missing something?

Considering Cauchy slices is an approach which can be used to prove the following and more general statement: if $(a_n)$ is a strictly positive sequence such that $\sum a_n$ diverges, denoting $S_n = \sum \limits_{k=1}^n a_k$, $\sum \frac{a_n}{S_n}$ also diverges.

Proof: let us bound below a Cauchy slice between $m$ and $n > m$:
$$\sum \limits_{k=m+1}^n \frac{a_k}{S_k} \ge \frac{1}{S_n} \sum \limits_{k=m+1}^n a_k = 1  - \frac{S_m}{S_n}.$$
Since $S_n$ goes to infinity, our lower bound goes to $1$ when $n$ goes to infinity, so the Cauchy slices' sums don't go to zero when $m \to \infty$.

In your problem, $a_n = \frac{1}{n}$, so denoting $H_n = \sum \limits_{k=1}^n \frac{1}{k}$, the previous remark about Cauchy slices ensures that $\sum \frac{1}{n H_n}$ diverges. 
Now all you need to do is explain why $H_n \sim \log(n)$, or at least why $\log(n) \le \mbox{Cste} \cdot H_n$, which you can do several different ways (including without integrals - see the comparison test on wiki).
