# Analytic continuation of a series raised to a power raised to a power?

## 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:

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

• Is the first line of the proof correct? – Szeto Oct 14 '18 at 23:35
• 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"). – Gottfried Helms Oct 15 '18 at 13:33

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 $$k=1$$.