$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function.

This series is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2< s< 1$ (convergence in this interval is essentially the Riemann hypothesis). Convergence at 1 is equivalent to prime number theorem.

Does divergence at 1/2 have similiar implications?

  • $\begingroup$ Accurately speaking, it is the analytic continuation, instead of convergence and divergence $\endgroup$
    – user40425
    Oct 25, 2013 at 7:25
  • 10
    $\begingroup$ Divergence at 1/2 means that the partial sums of the Mobius function cannot exhibit more than square-root cancelation. Or equivalently that the error term in the prime number theorem cannot be substantially smaller than $\sqrt{x}$. $\endgroup$
    – Lucia
    Oct 25, 2013 at 10:02
  • 5
    $\begingroup$ @Lucia: Perhaps you should post your comment as an answer, as it seems to resolve the question. $\endgroup$ Oct 27, 2013 at 20:15


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