Let $A$ be a set of positive integers and define $A_N:= \left\lvert A \cap \lbrace{ 1,2,\ldots,N \rbrace} \right\rvert.\ $ Is it true that $\displaystyle\sum_{n\in A} \frac{1}{n} $ diverges $\implies \displaystyle\sum_{N=1}^{\infty} \frac{A_N}{N^2}\ $ diverges? What about the reverse implication?
I think $\implies$ might be true based on the fact that the $N-$th prime number is $\approx \frac{N}{\log N}$ and the reciprocals of the primes diverges, and we also know that $\displaystyle\sum \frac{1}{N\log N}\ $ diverges.
Furthermore, I tried for a counter-example:
$$A= \bigcup_{n\in\mathbb{N}} \left[ \frac{2^{2^n}}{2}, 2^{2^n} \right] $$
because I know that $\displaystyle\sum_{n\in A} \frac{1}{n} \approx \sum_{n=1}^{\infty}\log 2,\ $ which diverges. However, $\displaystyle\sum_{N=1}^{\infty} \frac{A_N}{N^2} = \sum_{N=1}^{\infty} \frac{A_N/N}{N} > \left(\frac{1/3}{193} + \frac{1/3}{194} + \ldots + \frac{1/3}{256} \right) + \left(\frac{1/3}{49152} + \frac{1/3}{49153} + \ldots + \frac{1/3}{65536}\right) + \ldots \approx \frac{1}{3}\sum_{n=1}^{\infty}\log \frac{4}{3},\ $
which diverges. So this fails to be a counter-example.
Any thoughts or ideas on a counter-example or how to prove this?
Everything is positive, so we can interchange order of summation as we like. Write $$ \sum_{N=1}^\infty \frac{A_N}{N^2} = \sum_{N=1}^\infty\ \sum_{n \in A \cap \{1,...,N\}}\ \frac{1}{N^2}.$$
The double sum is over the set $$\{(N,n) : n \in A \cap \{1,...,N\}\} = \{(N,n) : n \in A , N \ge n\},$$ and so $$ \sum_{N=1}^\infty \frac{A_N}{N^2} = \sum_{n \in A} \ \sum_{N=n}^\infty \frac{1}{N^2}.$$ It remains to note that $$ \frac{1}{n} = \int_{n}^\infty \frac{1}{x^2}dx \le \sum_{N=n}^\infty \frac{1}{N^2} \le \int_{n-1}^\infty \frac{1}{x^2}dx = \frac{1}{n-1} \le \frac{2}{n} $$ (at least for $n \ge 2$, but for $n=1$ we have $\sum_{N=1}^\infty \frac{1}{N^2} = \frac{\pi^2}{6} \le \frac{2}{1}$, too).
Hence, $$ \sum_{n \in A} \frac{1}{n} \le \sum_{N=1}^\infty \frac{A_N}{N^2} \le 2 \sum_{n \in A} \frac{1}{n}.$$