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It is said that the Euler product $$\prod_p \frac{1}{1-p^{-s}}$$ diverges as $s \to 1^+$ proves we can't find constants $C$,$\theta$ with $\theta < 1$ such that $\pi(x) < C x^\theta$ because that would imply that the product converges for $s > \theta$.

I don't understand this deduction at all, how is the fact about the prime counting function concluded?

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up vote 8 down vote accepted

Taking logs of the Euler product, expanding out the resulting series, and comparing all the higher-order terms (i.e. terms involving $p^{-ns}$ for $n > 2$) to the convergent series $\sum_{m = 1}^{\infty} m^{-2s}$, we find that the divergence as $s \to 1^{+}$ is equivalent to the divergence of the series $\sum_p p^{-s}$ as $s \to 1^{+}$. Now if $\pi(x) < C x^{\theta}$, then the $n$th prime would have size $\sim n^{1/\theta}$, and so this series would behave like $\sum_{n = 1}^{\infty} n^{- s/\theta},$ which converges as $s \to 1^+$ (because $1/\theta > 1$), a contradiction.

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Matt, if I may ask, where'd the $m$ in the first sum come from? – J. M. May 16 '11 at 2:33
@J.M. Dear J.M., This is a typo, about to be corrected. Thanks! Regards, – Matt E May 16 '11 at 2:51
Matt, isn't the connection between pi(n) and the size of the n-th prime true only in average (so that the last sentence of the answer is correct, if read as an Abel summation or Tauberian argument)? – zyx Oct 5 '11 at 19:00
@zyx: Dear zyx, I didnt' think carefully about the argument for the last sentence; I think you are right that some kind of Tauberian argument might work, or perhaps if one takes into account a suffciently strong error term in the PNT? Best wishes, – Matt E Oct 6 '11 at 1:23

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