Let’s say we have an infinite set $A = \{a_i \mid a_i \in \mathbb{N} \}$ and such that $A$ is dense in $\mathbb{N}$ (in the sens that : $\lim_{n \to \infty} \frac{A \cap [[0,n]]}{n} \ne 0$), then is it possible that the following serie converges :
$$\sum_{a \in A} \frac{1}{a}$$
If the answer is yes, I am also interested about the more general question : Is it possible to find an infinit set $A$ dense in $\mathbb{N}$ such that $\forall \alpha \in (0,1]$, the série :
$$\sum_{a \in A} \frac{1}{a^{\alpha}}$$ converge.
So far I must say that I am enable to find such a set $A$. Moreover the « density » condition is extremely important because otherwise the problem is trivial because we can simply take $A$ to be the set of square numbers.
Thank you !
Write down the following partial sum: $$ S_{mn}=\sum_{a\in A\cap [0,mn]}\frac 1a=\sum_{k=1}^m\sum_{a\in A\cap[(k-1)n,kn]} \frac 1a\geq \sum_{k=2}^m\frac{|A\cap[(k-1)n,kn]|}{(k-1)n} $$ As $m\to\infty$, if the set $A$ is such that the cardinality $|A\cap[(k-1)n,kn]|$ is non-zero for sufficient number of times, then the series diverges.
Is the "dense" sets defined above one of those sets? Define: $$ c_k=|A\cap[(k-1)n,kn]|, $$ and note that using Stolz-Cesaro theorem: $$ \lim_{m\to\infty}\frac{\sum_{k=1}^mc_k}{mn}\neq 0\implies \lim_{k\to\infty}\frac{c_k}{n}\neq 0. $$
This means that for large enough $k$, $c_k$ is strictly non-zero and therefore the series diverges.