# Does the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ converge for $s>0$ as well as $\Re(s)>\sigma_c>0$?

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?

• The series diverges (and is unbounded) for $\sigma < \frac{1}{2}$. But to see that, you probably have to sum a lot more terms. $N = 10000$ is pretty small. Jan 9, 2021 at 20:32

$$\frac{\zeta(s+1)}{\zeta(s)}= \sum_{n=1}^\infty a_n n^{-s}$$ 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 $$\sup \{ \Re(\rho), |\Im(\rho)|\le T,\zeta(\rho)=0\}$$.
• Clarify if "The convergence on ${\rm Re}(s) = \sigma_0$" means "convergence on the whole line" or "convergence somewhere on the line". A necessary condition for a Dirichlet series $\sum c_n/n^s$ to converge at $s_0 = \sigma_0 + it_0$ where $\sigma_0 > 0$ is $(c_1 + c_2 + \cdots + c_N)/N^{\sigma_0} \to 0$ as $N \to \infty$. Therefore if (under RH) the Dirichlet series for $\zeta(s+1)/\zeta(s)$ is going to converge somewhere on the line ${\rm Re}(s) = 1/2$ we need $(\sum_{n \leq N} a_n)/\sqrt{N} \to 0$ as $N \to \infty$.
• @KCd Under the RH it can't converge anywhere on the line, because the convergence of $F$ at $s_0$ implies $F(s)=o(\frac{|s|}{\Re(s-s_0)})$ (last paragraph here). Honestly I'm not so sure about the converse, the coefficients of $\frac1{\zeta(s-1+\sigma_0)}$ are neither bounded nor non-negative so the Tauberian theorem relies a lot on $\frac1{\zeta(s)}=O(|\Im(s)|^\epsilon)$ on the region of analyticity. Jan 9, 2021 at 23:40