Holomorphy from polynomial growth of arithmetic partial sums (${\sum }_{n\le x}{a}_n=O(x^{\alpha })⟹\sum a_n{n}^{-s}\text{ holomorphic for }\mathrm{Re}s>\alpha$)

Let $S(x)={\sum }_{n\le x}{a}_n$ satisfy $|S(x)|\le C{x}^{\alpha }$ for $x\ge 1$. The Abel summation formula gives $A(s)=s{\int }_1^{\infty }S(t)t^{-s-1}dt$ when $\mathrm{Re}s>\alpha$, since its finite-sum boundary term tends to zero. Split this integral into blocks $I_m(s)={\int }_m^{m+1}S(t)t^{-s-1}dt$. For every $\delta >0$, the estimate $|I_m(s)|\le C{m}^{-1-\delta }$ holds uniformly on $\mathrm{Re}s>\alpha +\delta$. Each block is an entire function, by differentiation under the finite-range integral on compact sets. The Weierstrass M-test makes their sum locally uniformly convergent and therefore holomorphic; multiplication by $s$ preserves holomorphy.

This proof distinguishes uniform convergence of the block sum on a whole half-plane from local uniform convergence after multiplication by $s$. The latter is all that holomorphy needs. In particular, a bound ${\sum }_{n\le x}\mu (n)=O(x^{\alpha })$ gives a holomorphic continuation of the reciprocal Riemann zeta function to this half-plane and excludes zeta zeros there, with the pole at one treated by cancellation.