# 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).

• Does that mean that every serie that could be expressed by $\sum_{i=2}^n \frac{1}{n} an$ is divergent? – Daïshikofy Jun 2 at 14:47
• 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 – charmd Jun 3 at 11:24