Let $f(x) = \sum_{k\geq 0} a_kx^k$ be a real analytic function, converging on $-R < x < R$, whose $n^{th}$ Taylor polynomial is $f_n(x) = \sum_{k=0}^n a_kx^k$. If there exist $N\in\mathbb{N}$ and a sequence $\epsilon_n$ of values in $(0, R)$ such that $f_n > 0$ on $x\in(0, \epsilon_n)$ for $n\geq N$, then there exists $\epsilon\in(0,R)$ such that $f > 0$ on $x\in(0,\epsilon)$.
Is this statement true or false? If true, can it be strengthened? If false, is a weaker version true?
Edit: the statement is true provided that no subsequence of $\epsilon_n$ has limit 0. Certainly $f(x)\geq 0$ on $x\in\left(0,\liminf_{n\rightarrow\infty}\epsilon_n\right)$ by definition of convergence. Moreover, since $f$ is analytic, by the coincidence theorem for analytic functions there can be no sequence $x_n\rightarrow 0$ such that $f(x_n) = 0$; otherwise $f\equiv 0$, contradicting the hypothesis. So a suitable $\epsilon > 0$ must exist.
I still need to figure out what happens if a subsequence of $\epsilon_n$ approaches $0$.
The statement is true.
If $f(0) > 0$ then this follows by continuity. Otherwise $f(x)=x^kg(x)$ for some $k\in\mathbb{N}$ and real analytic $g(x)$ such that $g(0)\neq 0$. Since the lowest-order nonzero derivative of $f$ at $x = 0$ must be strictly positive, it follows that $g(0) > 0$, so that by continuity $f(x) > 0$ for small $x > 0$.