A number $a \in \mathbb{R}$ is transcendental if there isn't a polynomial $f(x) \in \mathbb{Q}[x]$ s.t. $f(a) = 0$.
I was thinking of generalizing this to infinite series. We will call a number $a \in \mathbb{R}$ series-trancendental if there isn't a power series $f(x) \in \mathbb{Q}[[x]]$ s.t. $f$ converges in $a$ and $f(a)=0$.
I was wondering if there is now a series-trancsendental number? (For example, $\pi$ isn't series-trancsendental because $\sin(\pi)=0$)
What if we expand the definition to allow $a \in \mathbb{C}$?
There are no series transcendental numbers. Given $a$ we can write a series with rational coefficients that has $f(a)=0$. Let $f(x)=\sum_{i=0}^\infty b_ix^i$ Choose $b_0 \in \Bbb Q$ such that $|b_0 -a| \lt1$ Now choose $b_1 \in \Bbb Q$ so that $|b_0+b_1a| \lt \frac 12$, then choose $b_2$ so that $|b_0+b_1a+b_2a^2| \lt \frac 14$ and so on. We have found a series such that $f(a)$ converges to $0$. Similarly if $a$ is complex, we can write $c=a\overline a$ and find a series in $x^2$ that takes $f(-c)$ to zero.