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Theorem Let $f : \mathbb{N} \to \mathbb{C}$ be an arithmetic function and let $M(f, x) = \sum_{n \leq x} f(n)$ be the summatory function of $f$. If $M(f, x) = Ax^{\alpha} + O(x^{\theta})$, where $\alpha > \theta \geq 0$, then the Dirichlet series $F(s) = \sum_{n = 1}^{\infty} f(n)n^{-s}$ has a meromorphic continuation to the half plane $\sigma > \theta$ with a simple pole at $s = \alpha$. More precisely, the function $F(s) - As(s-\alpha)^{-1}$ has analytic continuation to the half plane $\sigma > \theta$.

Let $\Lambda(n)$ be the von-Mangoldt function. If $$ M(\Lambda, x) = x + O_{\theta}(x^{\theta}), $$ where $1/2 < \theta < 1$ is fixed, how can I show that $\zeta(s)$ has no zeros in the region $\sigma > \theta$.

My Attempt: By the above theorem, I know that the Dirichlet series $$ \sum_{n = 1}^{\infty} \frac{\Lambda(n)}{n^s} = - \frac{\zeta'(s)}{\zeta(s)} $$ has a meromorphic continuation to the half plane $\sigma > \theta$. But I don't know how to proceed from here onwards.

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  • $\begingroup$ Since $\zeta(s)$ has a famous meromorphic continuation to $\mathbf C$, your right side $-\zeta'(s)/\zeta(s)$ is meromorphic on $\mathbf C$, not just on $\sigma > \theta$. The question you ask is treated in analytic number theory books, showing that if $\psi(x) = x + O(x^{\theta})$, then $\zeta(s) \not= 0$ for $\sigma > \theta$. See, for instance, Prop. 10.4 in Overholt's A Course in Analytic Number Theory. You need to use the Euler product for $\zeta(s)$, which is not something available with general Dirichlet series. The key phrase to look up in books is "Explicit formula". $\endgroup$
    – KCd
    Commented Dec 27, 2023 at 21:13
  • $\begingroup$ With the Explicit formula you can show that $\zeta(s) + \zeta'(s)/\zeta(s)$ is holomorphic on $\sigma > \theta$, so $\zeta'(s)/\zeta(s)$ has no zero when $\sigma > \theta$ (we already know $\zeta(s)$ is analytic in that half-plane except for a simple pole at $s = 1$). $\endgroup$
    – KCd
    Commented Dec 27, 2023 at 21:16

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If $\zeta(s)$ has a zero of order $m\ge1$ at $\rho$, then there exists some function $g(s)$, analytic and nonzero near $s=\rho$ such that

$$ \zeta(s)=(s-\rho)^mg(s), $$

so we have

$$ {\zeta'\over\zeta}(s)={m\over s-\rho}+{g'\over g}(s). $$

As a result, $(\zeta'/\zeta)(s)$ as a simple pole at $s=\rho$. If $\Re(\rho)>\theta$, then the meromorphic continuation of

$$ \sum_{n\ge1}{\Lambda(n)\over n^s} $$

in $\Re(s)>\theta$ will have a poles other than the simple pole at $s=1$, which contradicts the quoted theorem in the question.

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