The sum of natural numbers $ \sum_{n>0} x^n = \frac{x}{1-x}$, so as $x\to1^-$ it diverges as $(1-x)^{-1}$. So I wondered what would happen if we make the summation set thinner, i.e. $\sum_{n \in A} x^n$ for $A \subset \mathbb{N}$.
EDIT After receiving helpful suggestions, I guess I understand it a little better. If $A$ is a finite subset, than $ \lim_{x\to 1^-} \sum_{n \in A} x^n = \vert A \vert$. Thus
$ \lim_{x\to 1^-} \frac{\sum_{k \in (A \cap (1,n))} x^k}{\sum_{k=1}^n x^k} = \frac{\vert A \vert}{n} $
Taking the limit $n\to \infty$ the ratio gives the asymptotic density of $A$ in $\mathbb{N}$ if it exists. Indeed
$ \sum_{n=1}^\infty x^{2n} = \frac{x^2}{1-x^2} \;\;\; \text{ therefore } \;\;\; \lim_{x->1^-} (1-x) \frac{x^2}{1-x^2} = \frac{1}{2} $
What I still do not understand is how would I find the asymptotic behavior of the sum in case when the limit $\frac{\vert A\vert}{n}$ goes to zero, like in the case of $A$ being the set of prime numbers $A = \mathbb{P}$.
I guess the technique to be used here is to apply Mellin transform, and related the divergence rate to properties of zeta function.
$ \sum_{n>=1} \int_0^\infty t^{s-1} e^{-t n} dt = \Gamma(s) \sum_{n>=1} n^{-s} = \Gamma(s) \zeta(s) $
In the case of prime set this give $\Gamma(s) \zeta_\mathbb{P}(s)$ which goes as $log(s-1)$ for $s\to 1^+$, but I do not see it through yet..