All smooth functions have Taylor expansion? The theories are:

$f(x)$ has Taylor expansion equals the remainder of Taylor polynomial converge to $0$.

a smooth function such as $f(x) = e^{-1/x^2}$, $x&gt;0$, $f(x) = 0$, $x \leqslant 0$ do not has Taylor expansion near $0$.


However, I think about these propositions and find:
smooth - exist Taylor polynomial - select Peano remainder = $o(x^n$) - the remainder must converge to $0$ - smooth function must have Taylor expansion near $0$
I will be appreciated if someone can point out the error in this infer.
For every smooth function $f$, you can consider the Taylor series around $0$
$$\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$$
However, there are two possible obstructions in general to link directly this series to the function :

It is possible that the radius of convergence is $0$. Actually, one can prove that for every sequence $(a_n)$, there exists a smooth function $f$ satisfying $f^{(n)}(0)=a_n$. Therefore the Taylor series can be every series, so it can be a series which converges nowhere.

It is possible that the Taylor series converges everywhere, but is not equal to $f$ in a neighbourhood of $0$. The classical example is the one you mention with $e^{-x^2}$, you get a Taylor series which is the null series, and the function is non constant equal to $0$ in any neighbourhood of $0$.


Finally, you have the following condition for a function to be analytic over an interval $I$ : a smooth function $f : I \rightarrow \mathbb{R}$ is analytic over $I$ iff $\forall [a,b]\subset I$, there exists $M \in \mathbb{R}$ and $\alpha &gt; 0$ such that $\forall n \in \mathbb{N}$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$
Assume $f(x)$ as a smooth function and consider its $Maclaurin formula$,
then for every n:
$f(x) = \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k + o(x^n)$
select a sufficiently small neighborhood of zero, then:
$\lim\limits_{n\to+\infty}R_{n}(x) = \lim\limits_{n\to+\infty}(f(x) - \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k)=\lim\limits_{n\to+\infty}o(x^n)=0$
then smooth function $f(x)=\lim\limits_{n\to+\infty}\sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k=\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$.
then smooth function $f(x)$ has Taylor expansion.
the above conclusion is wrong, because $e^{-1/x^2}$ is smooth and do not have Taylor expansion. However, what's the error during my infer?