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I read in a text that Euler once proved the infinitude of primes by proving the divergence of their reciprocals, which seems to me a highly pleasing. However, I am only familiar with one proof of the primes' reciprocals' divergence and it requires the infinitude of the primes in one of its steps (it essentially compares the sum of primes cleverly with the harmonic series using some nice inequality manipulations).

I don't need to see Euler's original proof (which I am informed was insufficiently rigorous for my, modern, tastes anyway) but is there an elementary proof of the divergence of primes' reciprocals that does not require the infinitude of the primes?

Isky Mathews
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    One cannot prove the divergence of reciprocals of primes without the infinitude of primes, because it would be convergent for only finitely many primes. See here for a proof of the divergence. – Dietrich Burde May 02 '18 at 19:26
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    @DietrichBurde I think what Isky Mathews means is a proof of divergence of the reciprocals which does not rely on a prior proof of infinitude. Of course divergence of reciprocals would imply infinitude. – BallBoy May 02 '18 at 19:30
  • @Y.Forman: Thank you - this is precisely what I mean. – Isky Mathews May 02 '18 at 19:31
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    The proof does not rely on that, see here. – Dietrich Burde May 02 '18 at 19:31
  • I think (perhaps a modernized version of) Euler's proof had to do with the fact that the Zeta function has a pole at $1$, and $\log \zeta(s)$ can be identified with $\sum \frac1{p^s}$, so the fact that this diverges as $s \to 1$ proves the infinitude of the primes. I think this line of reasoning can be done rigorously. It relies on the Fundamental Theorem of Arithmetic, but not on infinitude of the primes. I don't have a reference on me at the moment. – BallBoy May 02 '18 at 19:33
  • Just saw Dietrich Burde's comment. Basically what Dietrich Burde linked to. – BallBoy May 02 '18 at 19:34
  • This may be relevant: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes#Euler's_proof – cansomeonehelpmeout May 02 '18 at 19:35

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Euler said.

$\zeta(s) = 1 + 2^{-s} + 3^{-s}+4^{-s} + \cdots\\ 2^{-s}\zeta(s) = 2^{-s} + 4^{-s} + 6^{-s} \cdots\\ (1-2^{-s})\zeta(s) = 1 + 3^{-s}+5^{-s} + \cdots\\ (1-2^{-s})(1-3^{-s})\zeta(s) = 1 + 5^{-s}+7^{-s} + 11^{-s} +\cdots$

This is effectively the sieve of Eranthoses

$\prod_\limits{p \text { prime}} (1-p^{-s})\zeta(s) = 1\\ \zeta(s) = \prod_\limits{p \text { prime}} \frac {p^s}{p^s-1}\\ \zeta(1) = 1 + \frac 12 + \frac 13 + \frac 14 + \cdots = \infty\\ \prod_\limits{p \text { prime}} \frac {p}{p-1} = \infty$

If the set of prime numbers were finite, then the left hand side would be finite, but as it isn't it ain't

Doug M
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