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I'm hoping to investigate the effects of divergent summation methods on series which cannot be analytically continued due to a dense set of singularities. At least a priori, it doesn't seem that a natural boundary should necessarily mean there aren't analytic functions that continue the original function. For example, we can easily generate somewhat contrived examples such as $\sum_{n=0}^\infty \frac{1}{2^{n-1}(n+1)^2}\sum_{k=0}^{2^{n-1}} (z-e^{\frac{2k+1}{2^n} \tau i})^{-1}$ This series absolutely converges whenever z is not exactly equal to a dyadic rational. On the other hand, it has a natural boundary at the unit circle because the dyadic rationals are dense on the unit circle.

So because the terms in this series eventually go to zero, the partial sums provide arbitrarily good approximations of the function when not around the unit circle, and so we have a natural way to "continue" this function to an analytic function using the partial sums.

We can easily craft many of these examples by enumerating the rationals--or some other dense set-- in steps and causing the later steps to eventually go to 0. Each of these functions will similarly have a natural boundary, yet can be approximated by analytic functions.

I'm interested in finding the validity of using divergent series methods to find continuations of functions with natural boundaries. For instance, one function I am looking at is $$\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{n(n+1)}{2}}}{\prod_{m=1}^{n}(1+x^{m})}$$ which is from one of Ramanujan's notebooks. It is special in that its terms grow quite slowly, so it's easier to handle with divergent series methods.

In particular, applying different types of smooth cutoff functions to this function leads to the same thing. Furthermore, the smooth cutoff method agrees with my approximation method. Graphed together, these methods look like this: enter image description here

The green line is the smooth cutoff function (using $\frac{1}{1.005^{n^2}}$), the orange line is the approximate method and blue is approximating Borel method (which ends up essentially like applying a smooth cutoff function). Each of these functions agrees with the starting function when the starting function converges.

Are there other ways to regularize this series that lead to this same result? Are there other methods that disagree? Also, because we don't have the identity theorem when we allow natural boundaries, is there a natural replacement of this theorem, possibly using properties of divergent series? My guess is that something like the Tauberian theorems could provide a good start for proving some sort of uniqueness, though it's not ideal because it requires somewhat strong restrictions on the properties of the terms in the summation.

I'd appreciate any insight into this problem, or insight into similar problems with faster divergences (for instance, I'm also interested in $\sum_{n=0}^\infty (-1)^n x^{n^2}$).

Finally, are there are any other proposed methods of continuing functions with natural boundaries that I can check the divergent series methods against? It's quite useful for me to be able to check my own results against the results of others to find where the methods agree and disagree, and hopefully why.

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    $\begingroup$ Are you familiar with Ecalle's Resurgence Theory? $\endgroup$ Sep 15 at 5:33
  • $\begingroup$ @PietroMajer I vaguely recognize the name. I will find some books and do some reading on the subject-- do you have any recommendations on the resources I should start with? $\endgroup$ Sep 15 at 5:39
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    $\begingroup$ David Sauzin has excellent notes available online, which I would recommend. But of you have a Google glance there may be something close to your specific interests $\endgroup$ Sep 15 at 5:55
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From those three examples, Rogers-Ramanujan's series belong to the class of basic hypergeometric series ($q$-series). It is a marginally logarithmic divergent series for $q > 1$ and it should be computed easily using traditional methods for alternating series. Just the third series, also belonging to this class, is really challenging and it deserves special attention. The following "simple" formal series

$$s(\zeta,q)= \sum_{n=0}^\infty (-\zeta)^n q^{n^2}$$

is closely related to Jacobi's $ϑ$ functions Fourier expansion (See either DLMF-NIST Handbook or Wikipedia). A proper choice of $ζ$ allows to get every $ϑ$ function. These $ϑ$ series (having strong convergence) are defined if nome $q = e^{iπτ}$ fulfills $0 < |q| < 1$ with $Im(τ) > 0$. As far as I know almost nothing is known off this range. For $q > 1$ the above series diverges far away factorial (Gevrey-$α$ series) divergence. This makes this kind of series really challenging and should be subject for a deep research. If you put $q^{n^{2}}$ as an exponential Taylor series in powers of $n^{2}\log(q)$ and swap the resulting sums you get a power series in $\log(q)$ whose coefficients are polylogarithms (See Addenda below). Except for some particular values of $\zeta$, the final series is still divergent (although much slower) with an additional issue, summands have an irregular sign pattern making difficult they can be fitted into some known summation method. Taking few terms of this asymptotic sum you can get about 2-3 decimal digits at most. In my humble opinion no currently known summation method can handle the above series (nor Jacobi's $ϑ$ Fourier series) in the general case when nome $q > 1$, however they can be computed using numerical quadratures as follows. By means of Nachbin's Theorem (see Wikipedia for a brief introduction), these series can be analytically continued (beyond Borel's Summation. See Ovidiu Costin - Asymptotics and Borel Summability) using this Mellin Transform

$$\int_0^\infty t^{s-1}\exp\left(-\frac{1}{a}\log^2(t)\right)dt = \sqrt{\pi a} e^{\frac{a}{4}s^2}, Re(a)>0$$

Lebesgue's dominance can be applied for $ζ ∈ ℂ$ \ $(-∞,0)$ to get $$s(\zeta,e^{a/4}) = \sum_{n=0}^\infty (-\zeta)^n e^{\frac{an^2}{4}} = \frac{1}{\sqrt{\pi a}} \int_0^\infty \frac{\exp(-\frac{1}{a}\log^2(t))}{1+\zeta t} \cdot \frac{dt}{t}$$

with $a = 4\log(q)$ and $|q|>1$, there are several interesting values for $ζ$ and $q$ to test, focused on Jacobi's thetas. For example $[ζ, q] → [ζ_{ } q^{±1/2},q^{±1/2}]$ with $q = 2$ or $q = e$. The multiprecision computation of this integral can provide as many digits as demanded. I hope this helps.

Addenda case $s(1,q) = \frac {1}{2}$ is proved using simple algebra $$s(\zeta,q) = 1 + \sum_{n=1}^\infty (-\zeta)^n e^{n^2\log(q)} = 1 + \sum_{n=1}^\infty (-\zeta)^n \sum_{k=0}^\infty \frac {n^{2k}\log^{k}(q)}{k!}$$ Swapping these sums we get

$$s(\zeta,q) = 1 + \sum_{k=0}^\infty \frac {\log^{k}(q)}{k!} Li_{-2k}(-\zeta)$$ where $Li_{-2k}(-\zeta)$ are polylogarithms of nonpositive even integer indexes. Using $Li_0(-\zeta) = (1 + \zeta)^{-1} - 1$ $$s(\zeta,q) = \frac {1}{1 + \zeta} + \sum_{k=1}^\infty \frac {\log^{k}(q)}{k!} Li_{-2k}(-\zeta).$$ But all summands vanish at $\zeta = 1$ since $$Li_{-k}(-\zeta)+(-1)^k Li_{-k}(-\zeta^{-1}) = 0$$ holds for all $ζ ∈ ℂ$ and $k = 1, 2, 3, ...$ (Erdélyi et al. 1981, § 1.11-17 pg 31).

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  • $\begingroup$ This is an amazing answer! I'll take a more thorough look at it as soon as I have time $\endgroup$ Sep 17 at 22:16
  • $\begingroup$ Am I interpreting your equation correctly? When I compute $$\sum_{n=0}^\infty (-1)^n x^{n^2}=\frac{1}{\sqrt{4\pi\ln(x)}}\int_{0}^{\infty}\frac{e^{\left(-\frac{\ln^{2}(t)}{4\ln(x)}\right)}}{t(1+t)}dt$$ I get 1/2 everywhere when x>1. Similarly, there are lots of angles in the complex plane where the function becomes a constant 1/2 when |z|>1. $\endgroup$ 2 days ago
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    $\begingroup$ @CelebBriggs. That's correct $s(1,x) = 1/2$. I will extend my answer to prove this. $\endgroup$ yesterday
  • $\begingroup$ I think I can see the s(1,x) argument pretty clearly, but I am more curious about the right way to extend this into the complex plane. I may have done something wrong, but allowing x to be complex doesn't quite work. Another consideration would be splitting the sum into parts, with each part being of the same form as the $s(1,x)$. So for the angle $\frac{1}{6} \tau$, I've split into $$\sum_{n=0}^\infty (-1)^n x^{(3n)^2} +e^{\frac{1}{6} \tau} \sum_{n=0}^\infty (-1)^n x^{(3n+1)^2} + e^{\frac{4}{6} \tau} \sum_{n=0}^\infty (-1)^n x^{(3n+2)^2}$$ $\endgroup$ yesterday

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