TITLE: If $A\subseteq\mathbb N$ and $\sum\limits_{a\in A}\frac1a$ converges then $A$ has natural density $0$
QUESTION [15 upvotes]: In this answer to a question about a series, a theorem was stated:

If $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the natural density of the set.

My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it?

REPLY [2 votes]: 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.