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This question already has an answer here:

Is it true that every sequence of real numbers $(a_n)$ can be a sequence of derivatives - ($f^{(n)}(0)$) of some function $f\in C^{\infty}(\mathbb{R})$?

It's clear if the series $\sum \frac{a_n}{n!}x^n$ has non-zero radius of convergency, but I didn't manage to prove it for the other cases.

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marked as duplicate by user127096, user127.0.0.1, John Habert, vonbrand, user61527 Mar 25 '14 at 17:49

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  • $\begingroup$ Don't you want infinitely differentiable just at $0$? $\endgroup$ – Pedro Tamaroff Feb 25 '14 at 17:37
  • $\begingroup$ I thought that there is no differnece, but your case is also interesting. $\endgroup$ – user122380 Feb 25 '14 at 17:40
  • $\begingroup$ I mean, I'm not saying only at $0$; but that that should suffice. $\endgroup$ – Pedro Tamaroff Feb 25 '14 at 17:40
  • $\begingroup$ See here: math.stackexchange.com/questions/63050/… $\endgroup$ – Etienne Feb 25 '14 at 17:42

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