# Divergence of the sum of the reciprocals of the primes

I know there are many ways to prove that the sum of the reciprocals of the primes diverges, but does the following argument work?

The cardinality of the set of all prime numbers is obviously ${\aleph_0}$. Intuitively we can map each natural number $N$ to a prime number ${p_N}$. Therefore $$\sum\limits_{n = 1}^\infty{\frac{1}{{p_n}}}$$ diverges because \begin{align} 1 &\mapsto {p_1}\\ 2 &\mapsto {p_2}\\ 3 &\mapsto {p_3}\\ &\vdots\\ \end{align} i.e. it resembles the harmonic series $$\sum\limits_{n = 1}^\infty{\frac{1}{n}}.$$ If it does not work, is there a way to make this argument work?

Edit: If this argument does not work in general, why does it make intuitive sense for the primes, but not for the reciprocals of the squares?

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Notice that for every $i,$ we have $\dfrac{1}{p_i}<\dfrac{1}{i},$ which doesn't help for the comparison test. – Andrew Dec 14 '12 at 0:27
The convergence of harmonic subseries is related to the density of its terms. See the work of Šalát et al; for instance eudml.org/doc/118201. – lhf Dec 14 '12 at 0:43
The question, "why does it make intuitive sense for the primes," is a question of psychology (specifically, a question of your psychology), not of mathematics. – Gerry Myerson Dec 14 '12 at 1:29
@GerryMyerson Almost all theorems are motivated by intuition and experience. We are not machines. – glebovg Dec 14 '12 at 1:32
True. But if that was meant as a reply to my comment, I don't see the connection. – Gerry Myerson Dec 14 '12 at 3:10

No, it doesn’t. If the argument worked, it would prove that $\sum_{n\ge 1}\frac1{n^2}$ diverges, since you could set up a similar correspondence.
If you want something fun to think about, consider the sum of the reciprocals of those positive integers with no $9$ in their decimal representations. What does your intuition tell you about them? – Gerry Myerson Dec 14 '12 at 1:31