3
$\begingroup$

I am interested if it is known, that is if was in the literature the following function $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ where $\mu(n)$ is the Möbius function and $z$ is the complex variable being $|z|<1$. See below my attempt and my motivation if you want to know why I am interested about it.

Question. For complex numbers $z$, we define the formal series $$ \Omega (z):=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ where $\mu(n)$ is the Möbius function. Does converge this complex function $\Omega (z)$ for $|z|<1$? How do you prove it? Was in the literature this function? If do you know some remarkable statements from a free-access source, please refer it. Many thanks.

Motivation. My calculations started from the generating function of Gregory coefficients, see in this Wikipedia. Then combining it with the Möbius inversion formula for the Taylor series of the logarithm and summation, if there are no mistakes I believe that I can state for $|z|<1$ $$\Omega(z)=z+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n G_n}{k}\Omega \left( z^{n+k} \right), $$ where $G_n$ is the $n$th Gregory coeffient.

My attempt. I know the series of the complex logarithm $Log(1-z)$ for $|z|<1$. I want (and I need it as a curiosity from my previous motivation) to justify the convergence of $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ for $|z|<1$. Then I tried the absolute convergence to state from $$ \left| \sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n \right| \leq\sum_{n=1}^{\infty}\frac{ \left| z \right|^n }{n}=-\log (1- \left| z \right| )$$ and since the RHS is convergent for our disk I believe that it justify the absolute convergence, this our genuine series is convergent on the open disk.$\square$

Thus I am asking, if my calculation/justification in the section attempt was right. Is appreciated as was asked if you tell me if the function $\Omega(z)$ for $|z|<1$ was in the literature, and references. If you know how justify different convergent issues for our function: uniform convergence or analyticity on the open disk, you can provide me the claims/references or hints.

$\endgroup$
1
  • $\begingroup$ If some user want (is interested in to know) more details to get the claim of the section Motivation I can add in next comment such details. Many thanks. $\endgroup$
    – user243301
    Feb 18 '17 at 20:17
2
$\begingroup$

As you proved, the series is converging absolutely and is thus converging, although here is a simpler proof: $$\left|\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n\right|\leq \sum_{n=1}^{\infty}\left|\frac{\mu(n)}{n}z^n\right|\leq \sum_{n=1}^{\infty}\left|z^n\right|=\frac{|z|}{1-|z|}$$ This series is also analytic within it's radius of convergence. Using Cauchy-Hadamard, the radius of convergence is $$R=\frac{1}{\limsup_{n\rightarrow \infty}\sqrt[n]{\left|\frac{\mu(n)}{n}\right|}}=\frac{1}{\lim_{n\rightarrow \infty}\sqrt[n]{\frac{1}{n}}}=1$$

$\endgroup$
1
  • $\begingroup$ Many thanks for your detailed answer, I need these details. I am waiting if some user know this function $\Omega(z)$ from the literature (because I've curiosity about this function), but your answer is perfect, and solve all my doubts. If in next weeks there are no more answers I will accept your answer. Many thanks for your attention and help. $\endgroup$
    – user243301
    Feb 19 '17 at 8:27

Your Answer

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