For which sequence do all associated power series converge? (describe the subset of $\Bbb R^\Bbb N$ "homeomorphic" to the analytic functions) I am investigating a certain property of real smooth functions (though it is easily extended to complex analytic functions) which requires me to define a topology on $C^\infty(\Bbb R)$ that "looks like" the standard topology on $\Bbb R^\Bbb N$. In my notes, I have written

Define the function $A:\Bbb R^\Bbb N\to C^\infty(\Bbb R)$ by
$$(A(\mathbf a))(x)=\sum_{n\in\Bbb N}a_nx^n$$
for all $\mathbf a\in\Bbb R^\Bbb N,x\in\Bbb R$. Let $\mathbf E^\Bbb N$ be the product topology on $\Bbb R^\Bbb N$, and define the topology $\tau_A$ on $C^\infty(\Bbb R)$ by $U\in\tau_A\iff A^{-1}(U)\in\mathbf E^\Bbb N$ (every open set in $\tau_A$ is the image of an open set in $\mathbf E^\Bbb N$ under $A$.)

(Yes, $0\in\Bbb N$, I will die on this hill!)
This is nonsensical in an obvious way: not every sequence of real numbers produces a smooth function, so $A$ is not well-defined, and the construction of the suggested topology doesn't make sense. This is such a minor point in context that I could simply dismiss it by waving my hands and saying "let $P$ be whatever subset of $\Bbb R^\Bbb N$ such that for each $\mathbf p\in P$ and $x\in\Bbb R$, the series $\sum_{n\in\Bbb N}p_n x^n$ converges." and replacing each subsequent occurrence of '$\Bbb R^\Bbb N$' with '$P$'. Nothing about the way the topology is defined would change - it is still fundamentally "the $\Bbb N$-dimensional Euclidean product topology imposed on $C^\infty(\Bbb R)$" - but I find it odd that I don't really have any idea what this set looks like. It feels like something I ought to know but never learned. It isn't addressed in any of the textbooks that I own and I haven't seen it any questions on this site. And now that I'm having to put actual effort into the problem, I'm realising that it might not be completely trivial, either.
For what subset of $\Bbb R^\Bbb N$ do the associated power series converge? Does it have any noteworthy topological properties as a subspace of $(\Bbb R^\Bbb N,\mathbf E^n)$? Are there any fundamental results associated with it?

Bounds:
Let $P$ be the set in question. Using the Taylor series, we know that for any $\mathbf p\in P$, there exists $a_n=\mathcal o(n!)$ such that $p_n=\frac{a_n}{n!}$ for all $n\in\Bbb N$. However, $\frac{o(n!)}{n!}$ is not included in $P$. Consider that $(n-1)!= o(n!)$. Since $\frac{(n-1)!}{n!}=\frac1n$, and $\sum_{n\in\Bbb N}\frac{x^n}n$ does not converge everywhere in $\Bbb R$, the sequence $\frac{n!}{(n+1)!}$ cannot belong to $P$. In this sense, $\frac{o(n!)}{n!}$ is an "upper bound" for $P$.
 A: The sequences of real (or even complex) numbers $(z_n)_{n\geqslant 0}$ such that for some positive real number $r$ the series $F(z) = \sum_{n\geqslant 0} z_n r^n$ converges are precisely the sequences of subgeometric growth: there must exist some $r_0$ and some natural number $n_0$ such that $|z_n|\leqslant r_0^n$ for all $n\geqslant n_0$. This is known as Abel's Lemma (or follows from it). Thus, your map is at least well defined on the set of sequences of subgeometric growth if what you want are germs of convergent power series.
On the other hand, if what you want are series that define a convergent power series on all the real line (equivalently, they define entire functions) then what you want (by the above) are those sequences that go to zero faster then any exponential sequence of the form $\varepsilon^n$ where $\varepsilon\in (0,1)$.
Proof of Abel's Lemma. If $F(z)$ converges for some $r>0$ then $|z_n|r^n$ converges to zero, so at some point we get the bound $|z_n|\leqslant r_0^n$ where $r_0r=1$. Conversely, if $|z_n|\leqslant r_0^n$ for $n\geqslant N$ then
$$\sum_{n\geqslant N} |z_n|s^n \leqslant \sum_{n\geqslant N} (r_0s)^n$$
and $F(z)$ converges absolutely when $|z|r_0<1$.
