Let $f\colon\mathbb{R}\longrightarrow\mathbb R$ , which is infinite time differentiable on $\mathbb R$. and $x_0$ $\in \mathbb{R}$ be any fix point.
Now, for $n\in\mathbb{N}$ and $x\in \mathbb{R}$ ,using Lagranges theorem on some specific function, we have:
$$ f(x)=f(x_0)+\dfrac{f'(x_0)}{1!} (x-x_0)+\cdots+\dfrac{f^{n}(x_0)}{n!}(x-x_0)^{n}+\dfrac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1}\text{ for some } c \in [x, x_0]\text{,} $$
where $c$ is depends on $n$, $f$, $x$ , $x_0$.
Let
$$ T_{n}(x)=f(x_0)+$\dfrac{f'(x_0)}{1!} (x-x_0)+\cdots \ldots+\dfrac{f^{n}(x_0)}{n!}(x-x_0)^{n} $$
and
$$ R_{n}(x)=\dfrac{f^{n+1}(c)}{(n+1)!}(x-x_0)^{n+1} \text{ for all } x \in \mathbb R\text{.} $$
Now, for $n\in\mathbb N$ and $x$ $\in \mathbb R$ have :
$$ \implies f(x)=T_{n}(x)+R_{n}(x)\text{,} $$
where $R_{n}(x)$ depends on $n$, $f$, $x$, $x_0$.
Now, from here I want to know what special property analytic function does have about $R_{n}(x)$.
What I am thinking is, if our $f$ is analytic at $x_0$ with radius of convergence $R'$, then
for any $x \in (x_0-R',x_0+R')$,
$$ \lim_{n\to\infty}R_{n}(x)=0\text{.} $$
Is my claim true? And what about converse, i.e., if $\lim_{n\to\infty}R_{n}(x)=0$ for all $x$ in some neighbourhood 0f $x_0$ then can we conclude that $f$ is analytic at $x_0$?