Non analytic numbers We know that some real numbers (actually, most of them) are not algebraic and the proof of this fact is beautiful: algebraic numbers, like polynomials with integer coefficients, are countable, contrary to $\mathbb R$, hence there should be some non algebraic numbers. 
I was wondering if any real number $\alpha$ is "analytic", in the sense that it exists a power series $\displaystyle{ \sum_{n\geq 0} a_n(x-a)^n }$ of positive radius, with $a_n\in{\mathbb Q}$, and $a_n\in{\mathbb Q}$ for any $n\geq 0$, such that $\alpha$ is a root of $\displaystyle{ \sum_{n\geq 0} a_n(x-a)^n }$.
Of course, the above proof for non algebraic number does not work, since power series are uncountable, but I would be surprised that the answer is no. In that case, is there a simple example and does it change something if we consider Laurent series instead of power series?
Thanks in advance.

Without loss of generality, we may assume $|\alpha| < 1$.
We can construct the desired Taylor series
$$ f(x) = \sum_{n=0}^{+\infty} a_n x^n $$ 
as follows. Let $f_k(x)$ be the polynomial
$$ f_k(x) = \sum_{n=0}^{k} a_n x^n $$


*

*Choose $a_0$ so that $0 < a_0 < \alpha$

*Choose $a_k$ so that $|f_{k-1}(\alpha) + a_k \alpha^k| < \alpha^{k+1} $ and $|a_k| < 1$.


The radius of convergence of the resulting series will be at least $1$, and since $|f_k(\alpha)| < \alpha^{k+1}$, we have $f(\alpha) = 0$.
