Can Taylor's expansion(finite number of terms) be always close enough to the original function?

Question

Suppose $f(x)$ is analytic on $[a, b]$, so we have $n$ terms of Taylor expansion $f_n(x)$ at $x=a$. Is this $$\lim_{n\to\infty}\sup_{x\in[a, b]}|f(x)-f_n(x)|=0$$ always true? I started by observing the reminder like: $$\frac{f^{(n+1)}(\xi)\cdot(x-x_0)^{n+1}}{(n+1)!}$$and $$\frac{1}{n!}f^{(n+1)}(\theta x)(1-\theta)^nx^{n+1}, 0\le\theta\le1$$or $$\frac{1}{n!}\int ^x_{x_0}f^{(n+1)}(t)(x-t)^ndt$$ but none of them tell me what will happen if $f^{n}(a)\to \infty$

Answer 1

An answer before the question was edited to mean the function is assumed to be (real)-analytic:

It is not true even without the supremum, because there exist smooth functions with divergent Taylor series.

If the function is real-analytic, then the standard theory guarantees that if the Taylor series around a point $a$ converges at some $x_0$, then it will converge uniformly in the set of points $x$ such that $|x-a|<|x-x_0|$. This follows immediately from Weierstrass' M -test. In particular, if the Taylor series converges at $b$, then it will converge uniformly in the set $|x-a|<|b-a|$.