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Suppose that $1/2+it$ is not a zero of the Riemann zeta function $\zeta$, where $t \in \mathbb{R}$. Can $1/\zeta(1/2+it)$ be expressed as a Dirichlet series ?

GH from MO
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Q_p
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    I think what you really want to ask is whether the Dirichlet series $\sum\mu(n)/n^{s}$ converges on $\Re(s)=1/2$ except at the zeros of $\zeta(s)$. Note that any number $c$ can be expressed by many Dirichlet series (e.g. by the constant series $c$). – GH from MO Jul 21 '18 at 12:37
  • @GHfromMO, thanks for the correction. So does $\sum_{n=1}^{\infty} \mu(n)n^{-s} $ converge at $\Re(s)=1/2$ when $\zeta(s)\neq 0$ ? – Q_p Jul 21 '18 at 13:00
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    (Almost) duplicate of this question. – Jarek Kuben Jul 21 '18 at 13:28
  • @JarekKuben, i didn't know of that question. Indeed, Lucia's answer seems to also address my question. That is, the series $\sum \mu(n)n^{-s}$ seems to be divergent at every point on $\Re(s)=1/2$. – Q_p Jul 21 '18 at 13:57

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As Jarek Kuben remarked, the question is almost a duplicate of this question. Lucia's response there can be adapted here. Briefly, for a fixed $t\in\mathbb{R}$, let us write $M_0(x):=\sum_{n\leq x}\mu(n)/n^{1/2+it}$. If $\lim_{x\to\infty}M_0(x)$ exists, the Riemann hypothesis and the identity $$\int_0^{\infty} sM_0(e^x)e^{-sx} dx = \frac{1}{\zeta(1/2+it+s)},\qquad\Re(s)>0,$$ hold true. Now, fixing $\Im(s)\neq 0$ and letting $\Re(s)\to 0+$, the right hand side can blow up at the rate of $1/\Re(s)$, while the left hand side is $o(1)/\Re(s)$. This is a contradiction, hence $\sum_{n=1}^\infty\mu(n)/n^{1/2+it}$ diverges for any $t\in\mathbb{R}$.

GH from MO
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