1
$\begingroup$

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:

$$ \lim_{ x \to 1 }(1-x)(b_1 x + b_2 x^{(2)^\kappa} + b_3 x^{(3)^\kappa} + \dots) = (\sum_{\tilde r = 1}^\infty \frac{ b_\tilde r }{\tilde r ^\kappa}) \frac{1}{\zeta(\kappa)} $$

$\endgroup$
  • $\begingroup$ Is the first line of the proof correct? $\endgroup$ – Szeto Oct 14 '18 at 23:35
  • $\begingroup$ The first line of the proof is incorrect for any $\kappa \ne 1$. for larger $\kappa$ this defines a "lacunary" series and no nice closed forms are available for such (A keyword for further search might be "transseries"). $\endgroup$ – Gottfried Helms Oct 15 '18 at 13:33
1
$\begingroup$

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 $k=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.