My goal is to prove (or disprove) that sufficiently smooth and quickly decaying cutoff functions being tacked on to a Taylor series correctly extend the radius of convergence to the analytic continuation. So, I'd like to show that $$\lim_{\varepsilon \to 0}\sum_{n=0}^{G(\varepsilon)} \eta(n,\varepsilon)a_nx^n$$ is equal to the analytic continuation. In general, it seems that increasing the number of terms in the sum widens the interval at which the polynomial approximates to the continuation. For instance, the green function approximates the blue function in the interval from (-1.7,0], but after adding another 10 terms the red function can approximate the interval (-2.8,0] Making the size of $\varepsilon$ smaller decreases the interval at which the approximation is valid but increases the accuracy of the approximation. So $G(\varepsilon)$ must be some function that grows fast enough so that decreases in $\varepsilon$ is met with increases in the number of terms great enough to cause the interval of convergence to increases as the limit is approached.

Research done so far

So far, I have been able to generate one polynomial that is able to analytically continue any function with a non-zero radius of convergence. The function I obtained is $$\lim_{\varepsilon \to 0^{-+}}\sum_{n=0}^{\frac{R}{\varepsilon}}\frac{f^{(n)}(0)}{n!}\prod_{k=0}^{n-1}\left(x+\varepsilon k\right)$$ where $\varepsilon$ approaches from either the positive or negative direction depending on which direction the continuation should take place, and $R$ is the desired radius of convergence. This polynomial is essentially Euler's method applied with a seed value of the derivatives of the function at a point. It can be shown to converge to the analytical continuation by arguing that:

1. Recursively re-centering a Taylor series to a new point within their radius of convergence allows one to analytically continue a function. After an arbitrarily large number of applications of re-centering, the error between the function and this method of continuation can still be made arbitrarily small
2. Euler method approximates re-centering the polynomial from $c$ to $c+\varepsilon$. It approximates that process close enough so that even after stepping through Euler's method $\frac{R}{\varepsilon}$ times, the difference between the recursive re-centering method and Euler's method is arbitrarily small.

So long as the polynomial does not step directly into (or within $\varepsilon$ distance) of a singularity, each re-centering will converge and this method will correctly analytically continue the function in the direction it steps.

Interestingly, this method seems to converge towards a smooth cutoff function for any choice of $a_n$. For instance, here the dotted line is the aforementioned continuation method, and the drawn line is the function $$a_n{\left(1-\frac{\varepsilon}{2}\right)^{n^{2}}}$$

Demos and Numerical tests

Here is a Desmos example showing how the different methods work: https://www.desmos.com/calculator/snsv8l2cna

Here is how the series converges in the complex plane. The function on the left is $\frac{1}{1-x}$ and the function on the right uses the cutoff $\left(\frac{1}{1+\varepsilon}\right)^{(n^2)}$. As the number of terms becomes very large, it appears to converge on at least all $\mathfrak{R}(z)<1$, but over time seems to be able to gradually converge past that line.
This method also manages to continue functions with infinite singularities. For instance, this function has a singularity at each $(n \in \mathbb{N},i)$ and $(n \in \mathbb{N},-i)$. (Here the approximation is on the left instead of the right).

Potential explanations/leads

I have at least a few theories as to why this conjecture should be true. One view is that the Taylor series fails at a certain radius because that is precisely when all the infinitesimals/epsilons that have been ignored start to matter. For instance, in expanding out the original formula, one gets that: $$\lim_{\varepsilon \to 0^{-+}}\sum_{n=0}^{\frac{R}{\varepsilon}}\frac{f^{(n)}(0)}{n!}\prod_{k=0}^{n-1}\left(x+\varepsilon k\right)= \left(\sum_{K=1}^{\frac{R}{\varepsilon}}x^{K}\left(\sum_{a=K}^{\frac{R}{\varepsilon}}\frac{f^{(n)}(0)}{n!}\left(\varepsilon\right)^{\left(a-K\right)}S\left(K+1,a-1\right)\right)\right)+f\left(0\right)$$ where S(n,m) are Stirling numbers of the first kind. What is interesting here, is that if we were to simply set positive powers of to $\varepsilon$ 0 (i.e. $\text{let}, \varepsilon^\alpha = 0, \alpha > 0$), then one gets exactly the usual Taylor series. In fact, if one separates out the terms with positive powers of $\varepsilon$ from the regular Taylor series, one gets something that very rapidly converges to 0 within the radius of convergence, and explodes to an infinite value outside that radius of convergence.

Further, if one tries to keep the infinitesimals when taking the derivatives it seems to give another method to analytically continue functions (I haven't proved this one works yet, but it also has worked on all the functions I tested). If one defines $\frac{d}{dx}f(x) = \frac{f(x+\varepsilon) - f(x)}{\varepsilon}$ without the limit attached, then one can get 'infinitesimal' Taylor series as $$\lim_{\varepsilon \to 0-+} \sum_{n=0}^{\frac{R}{\varepsilon}}a_nx^{n}, \quad a_n = \left(\sum_{k=0}^{n}\frac{\left(-1\right)^{\left(n-k\right)}}{\left(n-k\right)!k!}\frac{f\left(k\varepsilon\right)}{\varepsilon^{n}}\right)$$ Where R is radius of convergence of the original Taylor series (we need this so that we aren't cheating and computing $f(x)$ past where we already know it converges since otherwise $f(x\varepsilon)$ could lie outside of our original radius of convergence). This Taylor-like series, also produces $a_n$ that is close to the function $\frac{1}{(1+\varepsilon)^{(n^2)}}$ as this image shows

Another closely related view is that Taylor series fail to converge because the tail of these sequence does not converge. Using the integral definition of the error for a Taylor series, one can show that the error between the Taylor series for $\frac{1}{1-x}$ at 0 for $k$ terms is $\frac{x^{\left(k+1\right)}}{\left(x-1\right)}$. But, the Taylor series for that part is $-\sum_{n=k+1}^\infty x^n$. This is interesting since it suggests that the error is really just the remaining number of terms. So $$E(x) = -\sum_{n=k+1}^\infty x^n, f(x) = \sum_{n=0}^{k} x^n - E(x), f(x)= \sum_{n=0}^{k} x^n + \sum_{n=k+1}^\infty x^n$$ Under this view, the reason that some Taylor series don't converge is because the tail of terms don't go to zero for any finite choice of k. With this in mind, the reason a smooth cutoff function would work is because it is able to well approximate the tail of the sequence. Here, one has to prove that a $\sum_{n=0}^{G(\varepsilon)}a_n\left(1-\eta(n,\varepsilon)\right)$ well approximates $\int_{0}^{x} \frac{f^{(k+1)}}{k!} (x-t)^k dt$

I've been stuck on this for a while, and I'd appreciate any answer or advice on future directions to continue researching this.