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?