Sequences for that $\sum_{n} \frac{1}{x_n}$ is divergent and $\sum_{n} \frac{1}{x_n \ln x_n}$ is convergent

Question

We will denote with $(x_n)$ a given sequence and we introduce the following two series.

$$S^* = \sum_{n} \frac{1}{x_n} \quad \text{and} \quad S_* = \sum_{n} \frac{1}{x_n \ln x_n}.$$

We know that if $(x_n)$ are for example the Fibonacci numbers greater then $1$, then $S^*$ and $S_*$ are convergent. If $(x_n)$ are the prime numbers then $S^*$ is divergent and $S_*$ is convergent. If $(x_n)$ are the natural numbers greater then $2$, then both series are divergent.

Question. How could we characterise the $(x_n)$ sequences, for that $S^*$ is divergent and $S_*$ is convergent. I would be also glad to see some reference in this topic. If we cannot characterise $(x_n)$ then is there any special property of such sequences?

A crude requirement for both to be convergent is $x_n\approx a^n$ for some $a>0$, which is something Fibonacci numbers enjoy.
If you trust the Erdos Turan conjecture, then $S^$ diverging implies $x_n$ has arbitrarily long arithmetic progressions. An affirmation of this is the Green-Tao theorem for prime numbers. So perhaps it would be worth studying when $S_$ diverges/converges if $x_n$ has arbitrarily long arithmetic progressions. This seems difficult... — Alex R., Nov 27 '14 at 20:14
@AlexR. Erdős also proved in 1935 that for any sequence where no term divides another $S_*$ is convergent. See details here. — user153012, Nov 27 '14 at 20:21
Ah, you want to characterize all sets which are large under $\displaystyle\frac1x$ and small under $\displaystyle\frac1{x \log x}$? :-) — Daniel R, Nov 27 '14 at 20:33
Your question is clear (and +1 btw), I was just making a half-humorous reference to one of my own questions. — Daniel R, Nov 27 '14 at 20:37

score 7 · Accepted Answer · answered Nov 27 '14 at 20:42

I don't think you're going to find any characterization that's simpler than your definition itself. The boundary between convergence and divergence is very delicate. For example, if $x_n = n( \log n )(\log\log n) (\log\log\log n)^\alpha$, then $S_*$ converges regardless of the value of $\alpha$, but $S^*$ converges when $\alpha>1$ and diverges when $\alpha\le1$.

Sequences for that $\sum_{n} \frac{1}{x_n}$ is divergent and $\sum_{n} \frac{1}{x_n \ln x_n}$ is convergent

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