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EDIT2: It seems that with a combination of this method and another method, it's possible to analytically continue functions past natural boundaries, in a way that seems quite natural. For instance, the blue is the analytical continuation of the function (in red) $$ \sum_{n=0}^\infty x^{(n^2)} $$ which by Fabry gap theorem has a natural boundary on the unit circle. What I find especially interesting about this continuation is that at $x=-1$ the function has derivatives that are 0 at all orders, and the continuation converges to those derivatives while not becoming the constant function.

EDIT: I've added another example: https://www.desmos.com/calculator/ntwi3h0ick This one is significant because it shows that this method works even when the derivatives are don't follow a nice curve. Here $a_n = n\sin(1.3n)$, which seems like it would make it more challenging to deal with the tail. The smooth cutoff function does still work here, but it does take a fair bit more terms to converge well.

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 increases the radius of convergence. Making the size of $\varepsilon$ smaller decreases the radius of convergence but increases the accuracy. 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.

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.