Why doesn't the Stone-Weierstrass theorem imply that every function has a power series expansion? I know that not every function has a power series expansion. 
Yet what I don't understand is that for every $C^{\infty}$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say : 
$$\forall x \in [a,b], f(x) = \lim_{n \to \infty}  \sum_{k = 0}^{\infty} a_{k,n}x^k$$
But then because it converges uniformly why can't I say that : 
$$\forall x \in [a,b], f(x) =   \sum_{k = 0}^{\infty} \lim_{n \to \infty} a_{k,n}x^k$$
And so $f$ has a power series expansion with coefficients: $\lim_{n \to \infty} a_{k,n}x^k$. 

I'm not sure what sense you can give to
$$
\lim_{n\to\infty}a_{k,n}
$$
There's nothing in the Stone-Weierstrass that can even hint to the existence of such limit.
The traditional example of a $C^\infty$ function that's not everywhere analytic well illustrates this. Try and find a polynomial approximation for
$$
f(x)=\begin{cases}
0 & x=0 \\[4px]
e^{-1/x^2} & x\ne0
\end{cases}
$$
and you'll realize that the exchange is not possible, because the limit under scrutiny cannot exist.
