We denote for integers $n\geq 1$, $\mu(n)$ the Möbius function, see its definition from this MathWorld. And we define informally


where $s$ is the complex variable. I don't know if this definition and next exercise were in the literature, or are interesting. If you know such definition refers the literature.

Question. Where can I say that is defined our function $f(s)$, and what about the convergence of $f(s)$ for regions of the complex plane, improving my simple deductions below? Many thanks.

I know that for $\Re s>1$ one has $$f(s)=\sum_{n=1}^\infty\frac{1}{n}\left(\frac{1}{\zeta(s)}-\sum_{k=1}^n\frac{\mu(k)}{k^s}\right).\tag{1}$$

Additionally from the definition and using the triangle inequality, one has that $$ \left| f(s) \right|\leq\sum_{n=1}^\infty\frac{1}{n} \left| \sum_{k=n+1}^\infty\frac{\mu(k)}{k^s}\right| . \tag{2} $$

and from here I know that there was in the literature upper bounds for $\left| \sum_{k=1}^n\frac{\mu(k)}{k^s} \right|$ but I don't know if it help here and if my approach is good.

You can add an answer both in the case that your answer depends on the solution of open problems, and also in the case where you want to provide unconditionally a result.

  • $\begingroup$ What is the abscissa of convergence, assuming the PNT, and assuming the RH ? $\endgroup$ – reuns Aug 2 '17 at 12:01
  • $\begingroup$ I know from the literature an equivalence to RH, but it is concerning the Dirichlet series $\sum_{k=1}^\infty\frac{\mu(k)}{k^s}$. Are you saying that is exactly the same problem? Many thanks @reuns $\endgroup$ – user243301 Aug 2 '17 at 12:05
  • $\begingroup$ The abscissa of convergence is the first page of books on $\zeta(s)$ and Dirichlet series. Iff $\sum_{n=1}^\infty \mu(n) n^{-s}$ converges for $\Re(s) > \sigma$ then $\sum_{n=N}^\infty \mu(n) n^{-s} = \mathcal{O}(N^{\sigma-s+\epsilon})$. $\endgroup$ – reuns Aug 2 '17 at 12:08
  • $\begingroup$ Okey then I am going to take your notes in my notebook and to delete this question. Many thanks @reuns $\endgroup$ – user243301 Aug 2 '17 at 12:09
  • $\begingroup$ Also did you hear of summation by parts ?.... $\sum_{n=1}^\infty a_n b_n = \lim_{N \to \infty} a_N (\sum_{n=1}^N b_n) + \sum_{n=1}^{N-1} (\sum_{m=1}^n b_m) (a_n-a_{n+1})$. Here $a_n-a_{n+1} = \frac{1}{n}, b_n = \mu(n) n^{-s}$. $\endgroup$ – reuns Aug 2 '17 at 12:09

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