Let $(a_n)_{n=0}^\infty$ be an arbitrary sequence in a real Banach space $X$. Does there exist a smooth function $f: \mathbb {R} \rightarrow X$ such that $f^{(n)}(0)=a_n$ for $n=0,1,2,\ldots $?
According to The Convenient Setting of Global Analysis (Kriegl and Michor), this is due to Wells (1973). The statement given is:
15.4. Borel's Theorem. [Wells, 1973]. Suppose a Banach space $E$ has $C^\infty_b$bump functions. Then every formal power series with coefficients in $L_{\text{sym}}^n(E;F)$ for another Banach space $F$ is the Taylorseries of a smooth mapping $E \to F$.
In this case, $E = \mathbb{R}$ so you have $C^\infty_b$bump functions.
References:

$\begingroup$ I strongly believe that the original question has an affirmative answer for ALL Banach spaces. I would try to use the closed range theorem for Frechet spaces as e.g. in the Functional Analysis book of Meise and Vogt. $\endgroup$ – Jochen Wengenroth Sep 3 '12 at 14:15
The result you are aiming at should follow from the scalar case by using tensor products since $\mathscr C^\infty(\mathbb R,X)= \mathscr C^\infty (\mathbb R) \tilde{\otimes} X$ and $X^{\mathbb N_0} = \mathbb R^{\mathbb N_0} \tilde{\otimes} X$. Because of the nuclearity of $\mathscr C^\infty (\mathbb R)$ and $\mathbb R^{\mathbb N_0}$ the tensor norm does not matter and tensorizing a surjective (hence open) continuous linear operator with the identity leads again to a surjection.

$\begingroup$ (Replying to your comment on my answer here so you get notified of it) By "the original question", do you mean "the question asked here" or the general question of applying Borel's theorem to all pairs of Banach spaces, so dropping the requirement (in Wells' theorem) that the source space has sufficiently nice bump functions? $\endgroup$ – Loop Space Sep 3 '12 at 17:50

$\begingroup$ @Andrew: Sorry, I just misread the theorem you quote in your answer (confusing the roles of $E$ and $F$). I don't know whether the generalized Borel theorem (which Kriegl/Michor attribute to Wells although I did not see the result in the quoted article) holds for all Banach spaces. $\endgroup$ – Jochen Wengenroth Sep 4 '12 at 6:46