Background
I recently realized I could construct the below formula:
$$ \lim_{ x \to 1 }(1-x)(\sum_{r=1}^\infty b_r x^{r^\kappa} ) = (\sum_{\tilde r = 1}^\infty \frac{ b_\tilde r }{\tilde r ^\kappa}) \frac{1}{\zeta(\kappa)} $$
Is it correct? I was wondering if it already existed in the literature? Is it possible to write the series $\sum_{r=1}^\infty b_r x^{r^\kappa}$ be expressed as a Laurent series? If so what is the next term?
Proof
Consider the following expression:
$$ x^{s^\kappa} + x^{(2 s)^{\kappa}} + x^{(3s)^\kappa} + x^{(4s)^\kappa} + \dots = \frac{ x^{s^\kappa} }{1 - x^{s^\kappa}}$$
Let us write this in summation notation:
$$ \sum_{r=1}^\infty x^{(rs)^\kappa} = \frac{ x^{s^\kappa} }{1 - x^{s^\kappa}}$$
Let us multiply both sides with an arbitrary constant $a_\tilde r$:
$$ a_\tilde r \sum_{r=1}^\infty x^{(rs)^\kappa} = a_\tilde r \frac{ x^{s^\kappa} }{1 - x^{s^\kappa}}$$
Let us take $s \to \tilde r s$:
$$ a_\tilde r \sum_{r=1}^\infty x^{(r \tilde r) s^\kappa} = a_\tilde r \frac{ x^{(\tilde r s)^\kappa} }{1 - x^{(\tilde r s)^\kappa}}$$
Let us sum of $\tilde r$ from $1$ to $\infty$:
$$ \sum_{\tilde r=1}^\infty (a_\tilde r \sum_{r=1}^\infty x^{(r \tilde r s)^\kappa})= \sum_{\tilde r=1}^\infty a_\tilde r \frac{ x^{(\tilde r s)^\kappa} }{1 - x^{(\tilde r s)^\kappa}}$$
Now let us write the L.H.S sum in it's full glory! $$ \sum_{\tilde r = 1}^\infty a_\tilde r \frac{ x^{(\tilde r s)^\kappa} }{1 - x^{(\tilde r s)^\kappa}} =$$ $$ a_1 (x^{s^\kappa} + x^{(2 s)^{\kappa}} + x^{(3s)^\kappa} + x^{(4s)^\kappa} + \dots) $$ $$ +$$ $$ a_2 (0 + x^{(2 s)^{\kappa}} + 0 + x^{(4s)^\kappa} + 0 + \dots)$$ $$ +$$ $$ a_2 (0 + x^{(2 s)^{\kappa}} + 0 + x^{(4s)^\kappa} + 0 + \dots)$$ $$ +$$ $$ a_3 (0 + 0 + x^{(3s)^\kappa} + 0 + 0 + \dots) $$ $$\vdots$$ Vertically summing the terms and defining the coefficients $b_r$: $$ \underbrace{a_1}_{b_1} x^{s^\kappa} + \underbrace{(a_1 + a_2)}_{b_2} x^{(2s)^\kappa} + \underbrace{(a_1+a_3)}_{b_3} x^{(3s)^\kappa} + \dots = \sum_{\tilde r = 1}^\infty a_\tilde r \frac{ x^{\tilde r s^\kappa} }{1 - x^{\tilde r s^\kappa}} $$ Hence, $b_r = \sum_{k} a_k$ where the permissible values of $k$ are the factors of $r$ . Now let us take this expression and multiply both sides with $(1-x)$
$$ (1-x)(b_1 x^{s^\kappa} + b_2 x^{(2s)^\kappa} + b_3 x^{(3s)^\kappa} + \dots) = \sum_{\tilde r = 1}^\infty a_\tilde r \frac{ x^{\tilde r s^\kappa} }{1 - x^{\tilde r s^\kappa}} (1-x) $$
Taking limit both sides and using L' Hopital Rule for the R.H.S:
$$ \lim_{ x \to 1 }(1-x)(b_1 x^{s^\kappa} + b_2 x^{(2s)^\kappa} + b_3 x^{(3s)^\kappa} + \dots) = \lim_{ x \to 1 } \sum_{\tilde r = 1}^\infty \frac{ a_\tilde r }{(\tilde r s)^\kappa} $$
Let us now take $s \to 1$ both sides:
$$ \lim_{ x \to 1 }(1-x)(b_1 x + b_2 x^{(2)^\kappa} + b_3 x^{(3)^\kappa} + \dots) = \sum_{\tilde r = 1}^\infty \frac{ a_\tilde r }{\tilde r ^\kappa} $$
Now using the mobius inversion formula we have:
$$ \lim_{ x \to 1 }(1-x)(b_1 x + b_2 x^{(2)^\kappa} + b_3 x^{(3)^\kappa} + \dots) = (\sum_{\tilde r = 1}^\infty \frac{ b_\tilde r }{\tilde r ^\kappa}) \frac{1}{\zeta(\kappa)} $$
You wrote $\sum_{r=1}^\infty x^{(rs)^k} = \frac{ x^{s^k} }{1 - x^{s^k}} $.
But $\frac{ x^{s^k} }{1 - x^{s^k}} = \sum_{r=1}^\infty (x^{s^k})^r = \sum_{r=1}^\infty x^{rs^k} $ and $rs^k \ne (rs)^k =r^ks^k $ except for $k=1$.