Find a Dirichlet series for $\frac{\zeta(s-1)}{\zeta(s)}$ valid for $Re(s)>2$.

I know that we should use absolute convergence but not sure how that applies in this case.

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    $\begingroup$ The $Re(s)>2$ just follows simply from the half-plane of absolute convergence of the $\zeta$ function. Think about the abscissa of convergence of $\zeta(s)$ and what effect the transformation $s\mapsto s-1$ has. The import of the problem is finding a nice expression for the coefficients of the Dirichlet series. $\endgroup$ – sharding4 May 26 '17 at 22:27
  • $\begingroup$ Interestingly, $\displaystyle\frac1{\zeta(s)}=\prod_{k\ge1}e^{-\frac{P(kx)}k}$ where $P(x)$ is the prime zeta function. I suppose you could then expand this via Taylor expansion of $e^x$ and then multiply the entire product out... $\endgroup$ – Simply Beautiful Art May 26 '17 at 22:33
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    $\begingroup$ @SimplyBeautifulArt OP needs a Dirichlet Series. More likely needs $\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^{-s}}$ where $\mu(n)$ is the moebius function. $\endgroup$ – sharding4 May 26 '17 at 22:37
  • $\begingroup$ @sharding4 Yeah.... wait... hm... Yup, guess that solve everything $\endgroup$ – Simply Beautiful Art May 26 '17 at 22:38

If the Dirichlet series of $f$ and $g$ converge absolutely for some $s$, then $\text{D}(f,s)\cdot \text{D}(g,s)=\text{D}(f*g,s)$.

We know that $\text{D}(\mu,s)=\frac1{\zeta(s)}$ so it suffices to find a function whose dirichlet series is $\zeta(s-1)$; one easily sees that it is the function $N(n)=n$.

Hence $\frac{\zeta(s-1)}{\zeta(s)}=\text{D}(N*\mu,s)$ for $\text{Re}(s)>2$.

Now we use Möbius inversion to find out what $N*\mu$ is. Since $N*\mu=f$ iff $N=f*u$, where $u$ is the constant $1$ function, we seek for an $f$ which satisfy $\sum\limits_{d|n}f(d)=n$. This is satisfied by Euler's totient function $\varphi$, so we get that $\frac{\zeta(s-1)}{\zeta(s)}=\text{D}(\varphi,s)$.

  • $\begingroup$ You mean if the two Dirichlet series converge absolutely for $\Re(s) > \sigma$, then so does their product. $\endgroup$ – reuns May 27 '17 at 17:12
  • $\begingroup$ If the convergence is conditional, I don't think it is true, look at $\eta(s)^2$ whose abscissa of convergence should be related with the Dirichlet divisor problem $\endgroup$ – reuns May 27 '17 at 17:17
  • $\begingroup$ Right, I corrected it. Thanks. $\endgroup$ – Martin Čech May 28 '17 at 21:51

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