# Implications of divergence of $1/\zeta(s)$ at 1/2

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ 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?

• Accurately speaking, it is the analytic continuation, instead of convergence and divergence
– user40425
Oct 25, 2013 at 7:25
• 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 $\sqrt{x}$. Oct 25, 2013 at 10:02
• @Lucia: Perhaps you should post your comment as an answer, as it seems to resolve the question. Oct 27, 2013 at 20:15