Background
I recently noticed the following:
$$ S(x)=\sum_{r=1}^\infty x^{p_r} $$
where $p_r$ is the $r$'th prime:
$$ \sum_{r=1}^\infty S(x^r) = \sum_{r=1}^\infty \frac{x^{p_r}}{(1-x^{p_r})} $$
However, we also notice that:
$$ \sum_{r=1}^\infty S(x^r) = \sum_{r=1}^\infty \frac{x^{p_r}}{(1-x^{p_r})} = \sum_{r=1}^\infty \omega(r) x^r $$
Where $\omega(r)$ is the number of distinct prime factors of $r$ and assigning $\omega(1)=0$ : http://mathworld.wolfram.com/DistinctPrimeFactors.html (More about it here)
Using the mobious inversion formula:
$$ S(x)=\sum_{r=1}^\infty x^{p_r} = \sum_{r=1}^\infty \sum_{s=1}^\infty \mu(s) \omega(r) x^{rs}$$
We know
$$ \sum_{r=1}^\infty x^{p_r} \sim \frac{1}{(x-1)\ln(1-x)}$$
as $x \nearrow 1$
Questions
Is this correct?
Does this already exist? (If, so I will be more than happy for a reference)?
I was wondering if using this above information (including the link) can one say anything meaningful about $\mu(r)$ using this?
If not, what more information is required?
P.S: I have only taken one course in asymptotics in my physics degree (so basically I don't know much) ....