# Better bounds in the error term of the summatory function of Von-Mangoldt function and the Riemann Hypothesis

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.

• 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".
– KCd
Commented Dec 27, 2023 at 21:13
• 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 $s = 1$).
– KCd
Commented Dec 27, 2023 at 21:16

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.