Let $\{ a_n \}$ be a sequence of real numbers. Assume $g(x) = \sum a_k x^k$ converges on $(-r, r)$ for some $r>0$, then there exist $a, b >0$ and a neighbourhood $N \subset (-r,r)$ of $0$ such that "for any $n \in \mathbb N$ "
$$ \sup_{x \in N} |g^{(n)} (x) | \leq ab^n n!.$$
It is obvious that when $n = 0$, $g^{(n)}$ attains its maximum on a small closed interval $[-\delta, \delta].$ I think, when differentiated, the exponent goes down by one, so the factorial of about $n$ comes out. How can I prove it regolously? Please help me, thank you.
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I want to prove the following part
then there exist $a, b >0$ and a neighbourhood $N \subset (-r,r)$ of $0$ such that "for any $n \in \mathbb N$ "
$$ \sup_{x \in N} |g^{(n)} (x) | \leq ab^n n!.$$