Is there a sequence of rational numbers $a\_0, a\_1, \dotsc$ such that $\sum\limits\_{i\geq 0}a\_i x^i$ converges absolutely to $2^x$ for every $x\in \mathbb{Z}$?



Taylor series with coefficients in $\mathbb{Q}$
Yes.


You can construct such a series of the form


$$ 1 + \sum\_{i=0}^\infty (a\_i + b\_i x) \left( \frac{x^2}{(i+1)^2}\right)^{e\_i} \prod\_{j=-i}^i (x-j) $$


for some sequence $a\_i, b\_i$ of rational numbers and some sequence $e\_i$ of natural numbers (converging to $\infty$.)


This certainly defines a power series with rational coefficients since the $i$th term is divisible by $x^{2 e\_i}$ and $e\_i$ goes to $\infty$, so the coefficient of each power of $x$ is a finite sum of rational numbers.


The $\prod(x-j)$ term ensures that the $i$th term vanishes at every integer from $-i$ to $i$. Thus the function has the value $2^x$ at $x=i+1$ and $x=-i-1$ if and only if the sum of the first $i$ terms has the value $2^x$ at those points. For any values of $a\_1,\dots, a\_{i-1}, b\_1,\dots, b\_{i-1}, e\_1,\dots, e\_{i-1}$, there is a unique rational $a\_i,b\_i$ which ensures this power series takes the correct value at these $x$. (The denominator $i+1$ is so that the value of $e\_i$ does not affect this.)


Now we can choose $e\_i$ sufficiently large depending on $a\_1,\dots, a\_{i-1}, b\_1,\dots, b\_{i-1}, e\_1,\dots, e\_{i-1},a\_i,b\_i$ so that the version of the $i$th term with absolute values everywhere


$$(|a\_i| + |b\_i| |x|) \left( \frac{|x|^2}{(i+1)^2}\right)^{e\_i} \prod\_{j=-i}^i (|x|+|j|) $$
is as small as desired at the points $-i,\dots, i$. For example, we can ensure it is at most $2^{-i}$.


Having done this, the series is certainly absolutely convergent at all integer points (and thus at all points), because all but finitely many of the terms have absolute value at most $2^{-i}$ at any given point.


Choosing $a\_i,b\_i,e\_i$ according to this rubric, our series satisfies all the desiderata.


