Let $A$ be a subset of $\mathbb N$
Prove that $\sum_{n\in A} \frac{1}{n}$ converges $\implies \frac{1}{n} $ #$(A \cap[1,n]) \to_{\infty} 0 $
For the moment, I can see the reciprocal is false by taking the set $A$ of all primes. (The question was to study the link between the two statements).
Do you have an idea for that? Reasoning by contradiction does not seem easy to write.
Enumerate $A$ increasingly as $\{q_n, n\in \mathbb{N}\}$ (if $A$ is finite, there's no need to do this, because then the question is really easy).
Then the assumption is that $\displaystyle\sum \frac{1}{q_n}$ is a convergent series. But since its general term $\frac{1}{q_n}$ is decreasing, this implies $\frac{1}{q_n} = o(\frac{1}{n})$: $\frac{n}{q_n} \to 0$
Given $n$, pick $k$ such that $q_k \leq n < q_{k+1}$, then $\frac{1}{n} |(A\cap [1,n])| \leq \frac{1}{q_k} |(A\cap [1,q_k])| = \frac{k}{q_k} \to 0$ ($k$ tends to infinity when $n$ does)
Here's a cute proof. Let $c_n = |A\cap [1,n]|$ and $S_n=\sum_{a\in A\cap [1,n]}\frac 1a$. Our hypothesis is that $S_n$ converges.
Let $S_0=0$. Note that $$c_n = \sum_{k=1}^n 1_A(k)=\sum_{k=1}^n k\left(1_A(k)\frac 1k\right)=\sum_{k=1}^n k(S_k-S_{k-1}) = nS_n-\sum_{k=1}^{n-1}S_k$$
Thus $$\frac{c_n}n = S_n - \frac 1n \sum_{k=1}^{n-1}S_k$$
By Cesaro mean theorem, $\frac 1n \sum_{k=1}^{n-1}S_k$ converges to the same limit as $S_n$, hence $\lim_n \dfrac{c_n}n= 0$, which is what needed to be proved.
