Problem: Given an arbitrary sequence $a_n$ of positive numbers, is it possible to find some Schwarz function $f$ on $\mathbb R$ such that $$|f^{(n)}(0)| \geq a_n \, \forall n ?$$
If not, how fast can $b_n=f^{(n)}(0)$ grow?
Note: It is clear that the above is not true for analytic functions. But it is not clear to me if it could actually hold for Schwarz functions.
As r9m said the best answer is Borel's lemma saying the map sending smooth functions to their formal Taylor series at $0$ is surjective.
If there exists $(b_n)$ such that for all Schwartz function $f^{(n)}(0)/b_n \to 0$ let $$g_N(x)=\sum_{n=0}^N \frac{(-1)^n}{|b_{2n}| 2^n} \delta^{(2n)}(x)$$ then $$\lim_{n \to \infty} \langle g_n,\phi\rangle$$ converges for all Schwartz function $\phi$, it means so does $$\lim_{n \to \infty} \langle g_n,\hat{\phi}\rangle=\lim_{n \to \infty} \langle \hat{g_n},\phi\rangle=\langle \hat{g_\infty},\phi\rangle$$
Where $$\hat{g_\infty}(\xi)=\sum_{n=0}^\infty \frac{1}{|b_{2n}|2^n} (2\pi \xi)^{2n}$$ It is obviously wrong because $\hat{g_\infty}$ grows faster than polynomials so that $\frac1{1+\hat{g_\infty}}$ is Schwartz and clearly $$\langle \frac1{1+\hat{g_\infty}},\hat{g_\infty}\rangle= \infty$$