Does the Taylor series of an infinitely differentiable function converge; and if yes, does it converge to the function? I have googled it, but I am not satisfied with those.
So my questions are:

  
*
  
*Let $D$ be an open set in $\mathbb{R}$.
  
  
  Let $f:D\rightarrow \mathbb{R}$ be a infinitely differentiable function.
Fix $x_0\in D$
Then, does $\sum_{n=0}^\infty \frac{f^n(x_0)}{n!}(x-x_0)^n$ converge on some neightborhood of $x_0$?

Secondly,

  
*
  
*Let $D$ and $f$ be the domain and function illustrated as above.
  
  
  Fix $x_0 \in D$
Assume, $\sum_{n=0}^\infty \frac{f^n(x_0)}{n!}(x-x_0)^n$ converge on $W\cap D$ where $W$ is some neightborhood $W$ of $x_0$.
Then, does this series coincide with $f$ on $W\cap D$?

 A: 
Not every infinitely differentiable function possess a power series. For instance the function $ e^{-\frac{1}{x^2}} $ does not possess a power series centered at the point $x=0$.

A: Taylor's theorem says that every function $f(x)$ that has $n$ derivatives in some neighborhood of point $x_0$, is equal to the series (for every $x$ in that neighborhood)
$$f(x) = \sum_{n=0}^{n-1} \frac{f^n(x_0)}{n!}(x-x_0)^n + R_n(x)$$
Where $R_n$ is the remainder, 
$$
R_n(x) = \frac{(x-x_0)^{n}}{n!}f^{(n)}(\xi),  
$$
Where $\xi$ is some number between $x_0$ and $x$. Note that this remainder term is the just the generalization of the Lagrange mean value theorem, i.e. for n=1, 
$$
f(x) = f(x_0) + (x-x_0)f'(\xi)
$$
Now  if you let $n \to \infty$, the function will be equal to the infinite series (in a neighborhood of $x_0$)
$$
\sum_{n=0}^{\infty} \frac{f^n(x_0)}{n!}(x-x_0)^n
$$
Only if a neighborhood of point $x_0$ exists, such that for all x in that neighborhood the remainder term tends to zero
$$
\lim_{n \to \infty}R_n(x) = 0
$$
If this condition is satisfied the function is analytic at $x_0$.
You can easily prove that the remainder tends to zero if you can bound every derivative of your function, this is true for the sine function for example, for all $n$
$$
|\sin (x)^{(n)}| \leq 1 \Rightarrow \lim_{n \to \infty}\frac{(x-x_0)^{n}}{n!}|\sin (\xi)^{(n)}| \leq \lim_{n \to \infty}\frac{(x-x_0)^{n}}{n!} = 0
$$ 
Thus $\sin$ is an analytic function.
A: Borel's theorem (You can read about it here) provides strong counterexamples to your question.
