Given a sequence $\seq{a_k}$, is there a function $f$
having $f^{(k)}(0)=a_k$?

I.e., can you get any power series you want?
(in particular, ones with radius of convergence $0$)?

Yes:
consider
$$f(x):=\sum_{k=0}^\infty a_k x^k \phi_k$$
where $\phi_k$ is a very small bump function, depending on $a_k$.
EG, a bump function that is $\equiv 1$ on $[-\epsilon,+\epsilon]$
and $\equiv 0$ outside of $(-2\epsilon,+2\epsilon)$, where
$\epsilon \to 0$ really quickly.
IE, fast enough that the series actually converges;
it's enough if $\norm{a_k x^k \phi_k}_\infty < 2^{-k}$ (for $k>0$)
so say $\epsilon=(1/2a_k)^{1/k}$, or $(1/2)^{1/k}$ if $\abs{a_k}<1$.
This ensures that $f$ is defined; I may want to check that derivatives
work okay (at the origin, which is the only place where the sum is
infinite), so may wanna use $C^\infty$ metric instead?

You can even make the $\phi_k$ analytic, setting them to be
$e^{-Nx^2}$ for large enough $N$;
the trick is that analytic functions aren't closed in the uniform
metric -- or even in the $C^\infty$?
