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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?

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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).

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  • $\begingroup$ Does that mean that every serie that could be expressed by $\sum_{i=2}^n \frac{1}{n} an$ is divergent? $\endgroup$ – Daïshikofy Jun 2 at 14:47
  • $\begingroup$ No, the core idea is that if $\sum \frac{1}{n}$ diverges, then $\sum \Big(\frac{a_n}{\sum_{k=1}^n a_k}\Big)$ diverges $\endgroup$ – charmd Jun 3 at 11:24

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