= Holomorphy from polynomial growth of arithmetic partial sums
{title2=$\sum_{n\le x}a_n=O(x^\alpha)\ \Longrightarrow\ \sum a_nn^{-s}\text{ holomorphic for }\operatorname{Re}s>\alpha$}
Let $S(x)=\sum_{n\le x}a_n$ satisfy $|S(x)|\le Cx^\alpha$ for $x\ge1$. The <Abel summation formula> gives $A(s)=s\int_1^\infty S(t)t^{-s-1}dt$ when $\operatorname{Re}s>\alpha$, 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 $|I_m(s)|\le Cm^{-1-\delta}$ holds uniformly on $\operatorname{Re}s>\alpha+\delta$. 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.
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