1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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 $\frac{\zeta(s-1)}{\zeta(s)}=\text{D}(\varphi,s)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.