The notion of natural density gives a suficient condition. Namely, one can prove that if $A \subseteq \mathbb{N}$ has positive upper n. density then $\sum_{a \in A} \frac{1}{a}$ diverges. This condition is not a necessary one, though. A well-know well-known result in Number Theory ascertains that the series of the reciprocals of the prime numbers diverges whereas $\pi(n) = o(n)$.

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References

I. Erdös's brilliant proof of the divergence of $\sum_{p} \frac{1}{p}$: http://www.renyi.hu/~p_erdos/1938-13.pdf

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The notion of natural density gives a suficient condition. Namely, one can prove that if $A \subseteq \mathbb{N}$ has positive upper n. density then $\sum_{a \in A} \frac{1}{a}$ diverges. This condition is not necessary, though. A well-know result in Number Theory ascertains that the series of the reciprocals of the prime numbers diverges whereas $\pi(n) = o(n)$.

In the following list you are to encounter some previous discussions on MO that are closely related to the current inquiry of yours: