What's about $\sum_{n=1}^\infty\frac{\mu(n)}{n}r^n,$ when $0<r<1$?

I am interested in

Question. It is possible to get an expression or bounds (upper and lower bounds) for $$\sum_{n=1}^\infty\frac{\mu(n)}{n}r^n,$$ where $0<r<1$ is a fixed real, and $\mu(n)$ is the Möbius function?

The context of my question was the following (in fact I've deduced some expressions by specialization of the following statement, that is a Möbius inversion formula, for different arithmetic functions, and I would like edit some post about my questions), I've consider the specialization of the formula for Möbius inversion for Taylor series $$\frac{1}{p}=-\sum_{n=1}^\infty\frac{\mu(n)}{n}\log \left( 1- \frac{1}{p^n}\right)=\sum_{n=1}^\infty\frac{\mu(n)}{n}\sum_{m=1}^\infty\frac{1}{mp^{nm}},$$ considering $p$ and $p+2$ twin primes, thus I've written a similar expression for $p+2$ and after I take the sum of both statements, to deduce (symbolically) from Brun's theorem that $$\sum_{\text{twin primes}}\sum_{n=1}^\infty\frac{\mu(n)}{n}\sum_{m=1}^\infty\left(\frac{1}{mp^{nm}}+\frac{1}{m(p+2)^{nm}}\right)$$ is convergent.

In previous paragraph there was a typo, that I've fixed.

Now, with previous Question, I am asking if it is possible deduce an expression or bounds for $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\frac{1}{p^m}\right)^n$$ when $m$ is fixed (positive integer) and $p$ is also fixed (a fixed prime). Thanks in advance.

• The power series $\sum_{n=1}^{\infty} \frac{\mu(n)}n z^n$ has a radius of convergence $1$. Thus, it follows that the sum $\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\frac{1}{p^m}\right)^n$ converges for each $p$ and $m\geq 1$. – i707107 Aug 6 '16 at 1:06
• But, I am not sure if there exists any closed form for this. I doubt that it exists, because $\sum_{n=1}^{\infty} \frac{\mu(n)}n z^n$ is not a rational function in $z$. – i707107 Aug 6 '16 at 1:07
• Very thanks much for your comments, I don't know how prove that $\sum_{n=1}^{\infty} \frac{\mu(n)}n z^n$ is not a rational function in $z$, in any case I know that should be a difficult question, I say the related with the closed form, thus I am asking also by the bounds. Very thanks much one more time @i707107 – user243301 Aug 6 '16 at 5:39
• All users, I was interested in this question when I am considering some specializations as I've said, in an attempt to follow Glaisher calculations (page 77) in BENITO, NAVAS and VARONA, Möbius inversion from the point of view of flows, Proceedngs of the Segundas Jornadas de Teoría de Números, Biblioteca de la Revista Matemática Iberoamericana, here one has the article, from Universidad de La Rioja. – user243301 Aug 6 '16 at 5:52

I don't know if these bounds are sufficiently precise for you, but however it is a way. Since $\left|\mu\left(n\right)\right|\leq1$ we have, using the Taylor series of log $$\log\left(1-r\right)\leq\sum_{n\geq1}\frac{\mu\left(n\right)}{n}r^{n}\leq-\log\left(1-r\right)$$ and this is consistent to the fact that at $r=0$ the series is $0$, but it is not a good approximation when $r$ is near to $1$ since $$\sum_{n\geq1}\frac{\mu\left(n\right)}{n}=0\tag{1}$$ (maybe it is interesting to recall that $\sum_{n\geq1}\frac{\mu\left(n\right)}{n^{s}}=\frac{1}{\zeta\left(s\right)},\,\textrm{Re}\left(s\right)>1$). So we fix a positive integer $N$ and, by Abel's summation, we can see that $$\sum_{n\leq N}\frac{\mu\left(n\right)}{n}r^{n}=\left(\sum_{n\leq N}\frac{\mu\left(n\right)}{n}\right)r^{N}-\log\left(r\right)\int_{1}^{N}\sum_{n\leq t}\frac{\mu\left(n\right)}{n}r^{t}dt$$ and now taking $N\rightarrow\infty$ and using $(1)$ and the fact that $0<r<1$ $$\sum_{n\geq1}\frac{\mu\left(n\right)}{n}r^{n}=-\log\left(r\right)\int_{1}^{\infty}\sum_{n\leq t}\frac{\mu\left(n\right)}{n}r^{t}dt$$ but since $$\left|\sum_{n\leq t}\frac{\mu\left(n\right)}{n}\right|\leq1$$ we have $$-r\leq\sum_{n\geq1}\frac{\mu\left(n\right)}{n}r^{n}\leq r$$ which is a better approximation since $$e^{x}\leq \frac{1}{1-x}$$ if $0<x<1.$
• I believe that I've undestand all details, after summantion, integrate $\left| \log\left(r\right)\int_{1}^{\infty}\sum_{n\leq t}\frac{\mu\left(n\right)}{n}r^{t}dt\right| \leq r$, and you've used the definition of absolute value in your claim. Finally the observation $-\log(1-r)\geq r$ provide us the comparison between the two lower bounds that you've calculated, and by exponentiation and the inequality that you've said, shows that the second of such lower bounds is better than the first. The other calculations are nice also. It just to say that I believe that I understood your calculations. – user243301 Aug 7 '16 at 7:04