What's about $\sum_{n=1}^\infty\frac{\mu(n)}{n}r^n,$ when $0I 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.
 A: 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 $0<x<1.$
