= Reciprocal Dirichlet-series bound on boundary-zero multiplicity
Absolute convergence of $1/\zeta(s)=\sum\mu(n)n^{-s}$ for $\operatorname{Re}s>1$ gives $|1/\zeta(\sigma+it)|\le\zeta(\sigma)=O((\sigma-1)^{-1})$ as $\sigma\downarrow1$. If $\zeta$ is holomorphic at $1+it_0$, a zero of order $m$ there would give reciprocal growth of order $(\sigma-1)^{-m}$ on the horizontal approach. Thus $m\le1$. This proves a multiplicity restriction, not existence of such a zero or a zero-free-line theorem.
Back to article page
