Reciprocal Dirichlet-series bound on boundary-zero multiplicity

Absolute convergence of $1/\zeta (s)=\sum \mu (n)n^{-s}$ for $\mathrm{Re}s>1$ gives $|1/\zeta (\sigma +it)|\le \zeta (\sigma )=O((\sigma -1{)}^{-1})$ as $\sigma \downarrow 1$. If $\zeta$ is holomorphic at $1+i{t}_0$, a zero of order $m$ there would give reciprocal growth of order $(\sigma -1{)}^{-m}$ on the horizontal approach. Thus $m\le 1$. This proves a multiplicity restriction, not existence of such a zero or a zero-free-line theorem.