TITLE: Every power series is the Taylor series of some $C^{\infty}$ function
QUESTION [55 upvotes]: Do you have some reference to a proof of the so-called Borel theorem, i.e. every power series is the Taylor series of some $C^{\infty}$ function?

REPLY [22 votes]: Through this question, I was made aware of 

Ádám Besenyei. Peano's unnoticed proof of Borel's theorem, Amer. Math. Monthly 121 (2014), no. 1, 69–72.

In this short note, Besenyei presents a proof due to Peano of the theorem usually attributed to Borel. Peano's result first appeared in

Angelo Genocchi , Giuseppe Peano. Calculo differenziale e principii di calcolo integrale, Fratelli Bocca, Roma, 1884. 

Note that Borel's result first appeared in his dissertation, published as

Émile Borel. Sur quelques points de la théorie des fonctions, Ann. Sci. l’École Norm. Sup. (3) 12 (1895), 9–55. MR1508908.

Peano's proof is short, and completely different from Borel's. Besenyei provides full details. I present a sketch:
Given a sequence $(c_n)_{n\ge0}$ of real numbers, we want a $C^\infty$ function $f$ such that $f^{(n)}(0)=c_n$ for all $n$. Peano considers
 $$ f(x)=\sum_{k\ge0}\frac{a_k x^k}{1+b_kx^2}, $$
for $(a_n)_{n\ge0}$ arbitrary, and $(b_n)_{n\ge0}$ a sequence of positive numbers, chosen so that $f$ is indeed $C^\infty$ and can be differentiated term by term. Assuming that this is possible, one easily checks that $f^{(n)}(0)=a_n$ for $n=0,1$, and that if $n\ge2$, then
 $$ \frac{f^{(n)}(0)}{n!}=a_n+\sum_{j=1}^{\lfloor n/2\rfloor}(-1)^ja_{n-2j}{b_{n-2j}}^{j}. $$
To see the latter, consider the power series expansion of $\displaystyle \frac{a_k x^k}{1+b_kx^2}$, valid for $|b_kx^2|<1$, and note that it implies that its $n$-th derivative at $0$ is either $0$ (if $n-k$ is odd), or
 $$ n!(-1)^ja_{n-2j}{b_{n-2j}}^{j}, $$
if $n-k=2j$ for some $j$.
The point is that this recurrence allows us to define the $a_n$ (uniquely) in terms of the $b_n$ and the $c_n$, so that $f^{(n)}(0)=c_n$ for all $n$.
In order for the above to hold, one needs to ensure that $f$ so defined can indeed be differentiated term by term. For this, Besenyei checks that if $k\ge n+2$, then $(*)$
 $$\left|\left(\frac{a_kx^k}{1+b_kx^2}\right)^{(n)}\right|\le(n+1)!\frac{|a_k|k!}{b_k}|x|^{k-n-2}$$ so, if $b_k$ grows sufficiently fast with respect to $a_k$, then 
 $$ \sum_{k\ge n+2}\left|\left(\frac{a_kx^k}{1+b_kx^2}\right)^{(n)}\right| $$ is uniformly convergent on any finite interval, and the Weierstrass M-test allows us to differentiate termwise.
Finally, Besenyei proves $(*)$ in a straightforward fashion with estimates coming from Leibniz rule, after rewriting
 $$\frac{a_k x^k}{1+b_kx^2}=\frac{a_k}{b_k}\cdot\frac{x^{k-1}}2\left(\frac1{x+\frac1{\sqrt{b_k}}i}+\frac1{x-\frac1{\sqrt{b_k}}i}\right). $$