Power series for functions that have $e$ as a root

Question

Define an 'algebraic sympathizer' as any transcendental number that is root of at least one function that when expressed as a power series, each of the coefficients are rational. That is, for all $n$ we have $a_n\in\Bbb{Q}$ and express the function: $$f(x)=\sum_{n=0}^\infty a_n x^n$$ Example: $\pi$ is a sympathizer. Because for $f(x)=\sin(x)$, the rational coefficients $a_n$ are easily generated: $$a_n = \begin{cases} 0, & \text{if $n$ is even} \cr \frac{1}{n!}(-1)^{\frac{n-1}{2}},&\text{if $n$ is odd} \end{cases}$$ Is $e$ a sympathizer? Are all computable numbers sympathizers?

Answer 1

This is more of an extended comment that got too long.

This problem is a bit ambiguous as it is easy to prove that given any real numbers $b_0, b_1,\dotsc,b_n,\dotsc$ such that infinitely many of them are non zero (hence all are non zero by renumbering) you can find non zero rational numbers $a_k$ such that $\sum a_kb_k=0$ (the series converges with value zero).

This is done by induction constructing $a_0,\dotsc,a_n$ such that $$|s_n|=\left|\sum_{k \le n}a_kb_k\right| \le 1/n^2$$ and then picking $a_{n+1}$ rational and non zero so that $a_{n+1}b_{n+1}$ is close enough to $s_n$ and with the right sign so $|s_{n+1}| \le 1/(n+1)^2$ which can be done by approximating $-s_n/b_{n+1}$ close enough by a non-zero rational (here we need of course that $b_n \ne 0$ for all $n$).

In particular, there are such $a_k$ such that $\sum a_kc^k=0$ for any real non-zero $c$ and of course the power series $\sum a_nz^n$ converges for $|z|<c$ and at $z=c$ by construction.

The better question in my opinion is if there is an entire function with the required property, or at least if there is a power series with radius of convergence strictly larger than $|c|$.