As an initial note, let me show by example what I mean by the terminology 'residue at $\infty$' I use in the title. I assume there is some standard terminology for this stuff, so I'd appreciate it if someone in the comments would let me know what it is officially called.


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This first example is motivated by studying the series $f(x) = \sum_{n=0}^\infty (-1)^n x^{2^n}$. One way we might choose to study this series is to study $g(n) = \frac{\csc(\pi n)}{2 i} x^{2^n}$ (for fixed $x$), since, with the proper contour, we have 
$$\int_C g(n) dn = f(x)$$

Here is an image of $g(n)$ for $x=\frac{1}{2}$. Here, hue corresponds to angle, and brightness corresponds to the modulus (with bright being large, and dark being small).
[![enter image description here][1]][1]
Interestingly, if we let $C$ be the contour drawn in grey, then the value of the integral is non-zero, despite the fact that there are no poles. This happens because of the growth of $g(n)$ in the region bounded by the contour. Therefore, I think of these contours as enclosing a different type of residue-- I refer to it as the residue at $\infty$.  

These residues actually play an important role when analyzing this series. For instance, one generally expects that the  $(\sum_{n=0}^\infty f(n) x^n) + \sum_{n=1}^\infty f(-n)x^{-n}$ adds up to the residues of $f(n)$. In this case, this means looking at $\sum_{n=0}^\infty (-1)^n x^{2^n} + \sum_{n=0}^\infty (-1)^n x^{2^{-n}}$. (The sum on the right doesn't quite converge, but it's easy to see that its Cesaro sum converges. Thus, it's being evaluated in the Cesaro means sense). When we look at
$$\sum_{n=0}^\infty (-1)^n x^{2^n} + \sum_{n=1}^\infty (-1)^n x^{2^{-n}}$$  
it turns out it is indeed precisely equal to the sums over all the residues at $\infty$, For instance, we could compute the value of that sum by summing adding up the contour integral of $g(n)$ over the two orange contours
[![enter image description here][2]][2]

If we transform $f(x)$ by using a power series expansion of $e^{\ln(x)2^n}$, then we obtain
$$f(x) = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{(-1)^n 2^{nk}\ln(x)^k}{k!} = \sum_{k=0}^\infty \frac{\ln(x)^k}{k!} \sum_{n=0}^\infty (-1)^n 2^{nk} "=" \sum_{k=0}^\infty \frac{\ln(x)^k}{k!} \frac{1}{1+2^k}$$

That final step is invalid, and it actually changes the value of the sum. However, notice that if we define $\widetilde g(n) = \frac{\ln(x)^n}{n!} \frac{1}{1+2^n} \frac{1}{e^{2 \pi i n}-1}$, then $\widetilde g(n)$ actually has some new poles, namely, for all $n = \frac{\pi i + 2\pi i k}{\ln(2)}$ then $1+2^n = 0$. $\widetilde g(n)$ has all of the residues (with the sign swapped) at $\infty$ show up as residues from the poles of $\frac{1}{1+2^n}$. In particular, the pole at $n = \frac{\pi i }{\ln(2)}$ evaluates to exactly the negative of the value of the integral over that grey contour

[![enter image description here][3]][3]

Likewise, the contour of $g(n)$ over $\frac{2 \pi i}{\ln(2)}  + \infty \to \frac{2 \pi i}{\ln(2)} \to  \frac{4 \pi i}{\ln(2)} \to \frac{4 \pi i}{\ln(2)} + \infty$ evaluates to the negative of the pole at $n = \frac{3 \pi i}{\ln(2)}$ for $\widetilde g(n)$. So, that method actually provides a way to move all the residues at $\infty$ into regular poles on the imaginary axis, and so it gives us an easy way to calculate them.


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## Question ##
I'm interested in finding a way to compute the 'residues at $\infty$' of the next iteration of this series, which is 
$$F(x) = \sum_{n=0}^\infty (-1)^n x^{2^{2^n}}$$ 
For instance, one of these residues at infinity is captured by the following contour in grey

[![enter image description here][4]][4]
The sum of all the residues at $\infty$ is given by 
$$\sum_{n=0}^\infty x^{2^{2^n}}(-1)^n + \sum_{n=1}^\infty x^{2^{2^{-n}}}(-1)^n$$
Where again we must take the second sum using Cesaro Summation. The hope would be that there is some way to transform these badly behaved residues at $\infty$ into regular poles, like the previous transformation does for $\sum_{n=0}^\infty (-1)^n x^{2^n}$

Note that the previous method doesn't work here. In paritcular, if we expand everything out as series, we end up getting something along the lines
$$\sum_{k=0}^{\infty}\frac{\ln\left(x\right)^{k}}{k!}f\left(\ln\left(2\right)k\right)$$  
Where $f(x) = \sum_{n=0}^\infty x^{2^n}$. However, finding the poles of $f$ would require extending it outside its natural boundary, which seems infeasible. I'm interested in any thoughts on how to approach this problem. 

#### Update 1 ####

I think a reasonable candidate for the poles is to rewrite the series as something of the form $\sum_{n=0}^\infty \frac{p(x,n)}{1+2^{2^n}}$. This seems to fit on where I would expect the poles to be, but I don't yet see any way to choose $p(x,n)$. 

  [1]: https://i.sstatic.net/9X9JV.jpg
  [2]: https://i.sstatic.net/Gu8yl.png
  [3]: https://i.sstatic.net/Fmm3B.jpg
  [4]: https://i.sstatic.net/Fe22q.png