This question pertains to the Fundamental Theorem of Dirichlet series which is stated on Wikipedia as follows (where $s=\sigma+i\,t$):
There are now three possibilities regarding the convergence of a Dirichlet series, i.e. it may converge for all, for none or for some values of s. In the latter case, there exist $\sigma_c$ such that the series is convergent for $\sigma>\sigma_c$ and divergent for $\sigma<\sigma_c$.
Assuming the Riemann Hypothesis (RH) I believe the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) below is valid for $\Re(s)>\frac{1}{2}$.
(1) $\quad\frac{\zeta(s+1)}{\zeta(s)}=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a_n\ n^{-s}\right),\quad \Re(s)>\frac{1}{2}\quad\text{(assuming RH)}\quad\text{where }a_n=\frac{1}{n}\sum\limits_{d|n} d\ \mu(d)$
I've noticed the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) above also seems to converge for $s>0$. Figure (1) below illustrates the reference function $\frac{\zeta(s+1)}{\zeta(s)}$ in blue and formula (1) above evaluated at $N=100$, $N=1000$, and $N=10000$ in orange, green. and red respectively.
Figure (1): Illustration of formula (1) for $\frac{\zeta(s+1)}{\zeta(s)}$ evaluated at $N=100, 1000, \text{and }10000$ (orange, green, and red)
Question: Does the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) above converge for $s>0$ as well as $\Re(s)>\frac{1}{2}$ (assuming RH), and if so is this inconsistent with the Fundamental Theorem of Dirichlet series?
Best Answer
$$\frac{\zeta(s+1)}{\zeta(s)}= \sum_{n=1}^\infty a_n n^{-s}$$ If it converges at $s_0$ then it converges and is analytic for $\Re(s) >\Re(s_0)$. Whence $\Re(s_0)\ge \sigma_0$ where $\sigma_0=\sup \{ \Re(\rho), \zeta(\rho)=0\}$.
Conversely the main Tauberian theorem plus several growth estimates say that $$\sum_{k\le K} \mu(k) = O(K^{\sigma_0+\epsilon})$$ So that $$\sum_{n\le N} a_n = \sum_{d\le N} d^{-1} \sum_{k\le N/d} \mu(k)= \sum_{d\le N} d^{-1} O((N/d)^{\sigma_0+\epsilon})=O(N^{\sigma_0+\epsilon})$$ And hence $\sum_{n\ge 1}a_nn^{-s}$ converges for $\Re(s)>\sigma_0$.
The RH is $\sigma_0=1/2$. The PNT doesn't imply that $\sigma_0<1$, it can very well be $\sigma_0=1$. The convergence on $\Re(s)=\sigma_0$ is trickier. If some zero has real part $\sigma_0$ then it diverges everywhere on $\Re(s)=\sigma_0$. If no zero has real part $\sigma_0$ then it might depend on the rate of convergence of $\sup \{ \Re(\rho), |\Im(\rho)|\le T,\zeta(\rho)=0\}$.