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 ? 



On the Dirichlet series for $1/\zeta(s)$ at $\Re(s)=1/2$
As Jarek Kuben remarked, the question is almost a duplicate of [this question](https://mathoverflow.net/questions/164874/is-it-possible-to-show-that-sum-n-1-infty-frac-mun-sqrtn-diverg). 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 $t\in\mathbb{R}$.


