1
$\begingroup$

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.

Illustration of formula (1) evaluated at N=100, 1000, and 10000 (orange, green, and red)

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Daniel Fischer
    Jan 9, 2021 at 20:32

1 Answer 1

Reset to default
1
$\begingroup$

$$\frac{\zeta(s+1)}{\zeta(s)}= \sum_{n=1}^\infty a_n n^{-s}$$ 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 $\sup \{ \Re(\rho), |\Im(\rho)|\le T,\zeta(\rho)=0\}$.

$\endgroup$
2
  • $\begingroup$ 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 $N \to \infty$. $\endgroup$
    – KCd
    Jan 9, 2021 at 23:22
  • $\begingroup$ @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. $\endgroup$
    – reuns
    Jan 9, 2021 at 23:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .