I have heard that the set of all functions $R \to R$ is more than uncountable, but the set of all continuous functions $R \to R$ is only uncountable.
If I reason correctly, the set of all Maclaurin series is uncountable, since the set of possible coefficients for each monomial is uncountable, while the set of monomials is countable, so the set of all possible polynomials is uncountable (the Cartesian product of a countable and uncountable set is uncountable).
However, not every continuous function is representable in Maclaurin series (any polynomial is C-infinite, while the set of all continuous functions is C-0). My question is: is every C-infinite function representable in Maclaurin series? If not, what's the name of the set of all functions $R \to R$ representable in Maclaurin series?
In order to put an end to this question (see this post as well on the issue), I just post my previous answer from the comments section; also feel free to ask further questions.
The standard example of a smooth function that cannot be represented by a Maclaurin series is
$$ f(x) = \begin{cases} e^{-1/x} & x>0, \\ 0 & x \leq 0 \end{cases}.$$
It is provable that $f$ is smooth at $x=0$, and it's also clearly smooth elsewhere. But its Maclaurin series is the constant zero function, which is not equal to $f$. The class of functions representable at $p$ by the Taylor series about $p$ are called real-analytic functions at $p$. And so the class of real analytic functions is strictly smaller than that of smooth functions.