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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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    True. But if that was meant as a reply to my comment, I don't see the connection.2012-12-14

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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.

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    @Gerry: He seems to have given the first published proof of its convergence; see [the Wikipedia article](http://en.wikipedia.org/wiki/Kempner_series). MathWorld uses the name, for whatever that’s worth.2012-12-14