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EDIT: I was considering some ways to get exact equations for the partial sums of some of those 0 radii of convergence functions. One way we can get the partial sums of the geometric series is to notice that $$S_n = \sum_{k=0}^n x^k \implies xS_n - S_n = x^{n+1}-1 \implies S_n = \frac{x^{n+1}-1}{x-1}$$ However, since we have removed the summation, $S_n$ depends on something continuous and so we might expect $S_n$ to agree with the calculated partial sum even at non-integer values.

Since we already have a differential equation representation for many of our sums, it struck me that we can do something similar to obtain exact forms for the partial sum formula. In the case of $S_N = \sum_{n=1}^N (n!)^2 (n+1)x^{-n}$, we have the differential equation $$\frac{d^2}{dx^2}S_N - \frac{1}{x}S_N + \frac{1}{x^2} = (N!)^2 (N+1) x^{-(N+2)}$$

Letting Mathematica solve this gives something in terms of the Meijer-G and Bessel functions. Interestingly though, this does very closely agree with the approximations of the partial sums. However, I think this strategy can't work in general, since finding the differential equation that the function solves is probably not possible for all functions. However, it still might be possible that there is an approximation method that finds approximate differential equations, though I don't have a good idea for how to do this.

As a final thought, I have usually been comparing the 'right' value to the value that other methods give, though a slightly more concrete requirement I often use is that if the series is a formal series to a differential equation, its regularization should solve the differential equation. For some of the series above, I have found the corresponding differential equations which can be transformed into the original equation: $$f'-2xf + 1 = 0 \to \sum_{n=0}^\infty \frac{(-1)^n}{2\sqrt{\pi}}\left(n-\frac{1}{2}\right)! x^{-(2n+1)} $$ $$f'' - \frac{1}{x}f + \frac{1}{x^2} = 0 \to \sum_{n=0}^\infty (n-1)!^2 n x^{-n} $$ $$f^{(k)} + (-1)^{k}f-\frac{(-1)^k}{x} = 0 \to \sum_{n=0}^\infty (kn)! (-1)^n x^{-(kn+1)}, k \in \mathbb{N}$$ $$ \frac{ds}{dx} = \frac{ (x-s)}{x^2} \to \sum_{n=0}^\infty (-1)^n n! x^{n+1}$$ (this last one is taken from https://math.stackexchange.com/questions/1832501/the-divergent-sum-of-alternating-factorials). Though, I suspect that thinking of divergent series as simply tools that solve differential equations is too narrow since I suspect there is some greater truth lurking beyond all of this.

As a final thought, I have usually been comparing the 'right' value to the value that other methods give, though a slightly more concrete requirement I often use is that if the series is a formal series to a differential equation, its regularization should solve the differential equation. For some of the series above, I have found the corresponding differential equations which can be transformed into the original equation: $$f'-2xf + 1 = 0 \to \sum_{n=0}^\infty \frac{(-1)^n}{2\sqrt{\pi}}\left(n-\frac{1}{2}\right)! x^{-(2n+1)} $$ $$f'' - \frac{1}{x}f + \frac{1}{x^2} = 0 \to \sum_{n=1}^\infty (n-1)!^2 n x^{-n} $$ $$f^{(k)} + (-1)^{k}f-\frac{(-1)^k}{x} = 0 \to \sum_{n=0}^\infty (kn)! (-1)^n x^{-(kn+1)}, k \in \mathbb{N}$$ $$ \frac{ds}{dx} = \frac{ (x-s)}{x^2} \to \sum_{n=0}^\infty (-1)^n n! x^{n+1}$$ (this last one is taken from https://math.stackexchange.com/questions/1832501/the-divergent-sum-of-alternating-factorials). Though, I suspect that thinking of divergent series as simply tools that solve differential equations is too narrow since I suspect there is some greater truth lurking beyond all of this.

As a final thought, I have usually been comparing the 'right' value to the value that other methods give, though a slightly more concrete requirement I often use is that if the series is a formal series to a differential equation, its regularization should solve the differential equation. For some of the series above, I have found the corresponding differential equations which can be transformed into the original equation: $$f'-2xf + 1 = 0 \to \sum_{n=0}^\infty \frac{(-1)^n}{2\sqrt{\pi}}\left(n-\frac{1}{2}\right)! x^{-(2n+1)} $$ $$f'' - \frac{1}{x}f + \frac{1}{x^2} = 0 \to \sum_{n=0}^\infty (n-1)!^2 n x^{-n} $$

 $$ \frac{ds}{dx} = \frac{ (x-s)}{x^2} \to \sum_{n=0}^\infty (-1)^n n! x^{n+1}$$ (this last one is taken from https://math.stackexchange.com/questions/1832501/the-divergent-sum-of-alternating-factorials). Though, I suspect that thinking of divergent series as simply tools that solve differential equations is too narrow since I suspect there is some greater truth lurking beyond all of this.

As a final thought, I have usually been comparing the 'right' value to the value that other methods give, though a slightly more concrete requirement I often use is that if the series is a formal series to a differential equation, its regularization should solve the differential equation. For some of the series above, I have found the corresponding differential equations which can be transformed into the original equation: $$f'-2xf + 1 = 0 \to \sum_{n=0}^\infty \frac{(-1)^n}{2\sqrt{\pi}}\left(n-\frac{1}{2}\right)! x^{-(2n+1)} $$ $$f'' - \frac{1}{x}f + \frac{1}{x^2} = 0 \to \sum_{n=0}^\infty (n-1)!^2 n x^{-n} $$ $$f^{(k)} + (-1)^{k}f-\frac{(-1)^k}{x} = 0 \to \sum_{n=0}^\infty (kn)! (-1)^n x^{-(kn+1)}, k \in \mathbb{N}$$ $$ \frac{ds}{dx} = \frac{ (x-s)}{x^2} \to \sum_{n=0}^\infty (-1)^n n! x^{n+1}$$ (this last one is taken from https://math.stackexchange.com/questions/1832501/the-divergent-sum-of-alternating-factorials). Though, I suspect that thinking of divergent series as simply tools that solve differential equations is too narrow since I suspect there is some greater truth lurking beyond all of this.