I am looking to create polynomials that converge past the usual radius of convergence on the real line. So far, I have proved that this polynomial will converge to the analytic continuation of the function a ray on the real line. R is the radius of convergence of the normal Taylor series. $$\lim_{\varepsilon \to 0^{-+}}\sum_{a=0}^{\frac{R}{\varepsilon}}\frac{f^{(n)}(0)}{n!}\prod_{k=0}^{a-1}\left(x+\varepsilon k\right) = \left(\sum_{K=1}^{\frac{R}{\varepsilon}}x^{K}\left(\sum_{a=K}^{\frac{R}{\varepsilon}}\frac{f^{(n)}(0)}{n!}\left(\varepsilon\right)^{\left(a-K\right)}S\left(K+1,a-1\right)\right)\right)+f\left(0\right)$$ where S(n,m) are Stirling numbers of the first kind. The side at which $\varepsilon$ approaches 0 depends on which direction the ray should face.
I would like to use the fact that I have one polynomial which converges to the analytic continuation of the function to prove other methods of analytic continuation work.
My goal is to prove at whether these methods work: $$\lim_{\varepsilon \to 0^{+}}\sum_{n=0}^{\left(1+\varepsilon\right)^{\frac{1}{\varepsilon^2}}} e^{-ln(1+\varepsilon)n^2}x^n $$ $$\lim_{\varepsilon \to 0-+} \sum_{n=0}^{\frac{R}{\varepsilon}}a_nx^{n}, \quad a_n = \left(\sum_{k=0}^{n}\frac{\left(-1\right)^{\left(n-k\right)}}{\left(n-k\right)!k!}\frac{f\left(k\varepsilon\right)}{\varepsilon^{n}}\right)=\sum_{k=0}^{n}\frac{\left(-1\right)^{\left(n-k\right)}}{\left(n-k\right)!k!}\frac{\sum_{m=0}^{\infty}\frac{f^{\left(m\right)}\left(0\right)}{m!}\left(k\varepsilon\right)^{m}}{\varepsilon^{n}}$$
The current difficulty I have is that it's very difficult to compare these polynomials term by term since it's difficult to gauge their behavior under the limit. Are there any tricks to proving that different polynomials converge to the same function? Do you have any tips for how I should approach this?