Radius of convergence of all taylor series of f are uniformly bounded then analytic Let $f \in C^{\infty}( \mathbb{R},\mathbb{R})$ such that the radius of convergence of all taylor series of $f$ are uniformly bounded . Then $f$ is analytic. 
I don't know if this result is true or not , i was working on analytic functions and i asked myself this question . I searched to find something similar , i am not able to prove it or find an example ...
 A: Edit (as noted by @Phoemue) - the argument is not correct since the first line is not correct and one has to be more subtle and use category arguments (Baire theorem about countable union of sets that exhaust an interval say) and Bernstein theorem about absolute monotonicity implying analyticity, so I will present the correct argument below and I apologize for being hasty)
****By hypothesis $|\frac{f^{(n)}(t)}{n!}| \le R^n$ for some fixed $R>0$ and all $t \in \mathbb R$**** is not correct
What is correct is that there is $n(t)$ for which the above happens when $n \ge n(t)$ and unfortunately there is no simple property of $n(t)$ here to allow to deduce that it is locally uniformly bounded since then the incorrect proof given earlier would still hold.
So one has to argue by contradiction and assume there is $x_0$ where $f$ is not analytic and note that $x_0$ cannot be isolated since otherwise, $f$ is analytic on $x_0 -\delta < x <x_0 \cup x_0<x< x_0+\delta$. This means there are analytic function $g_1, g_2$ which have the OP property (radius of convergence of Taylor series at least $1/R$) and equal to $f$ to the left and right of $x_0$ respectively; but with $\delta < 1/R$ it follows that $g_1$ extends through $x_0$ analytically and hence that $g_1^{(k)}(x_0)=f^{(k)}(x_0)$ and same for $g_2$ so $g_1=g_2=f$ on a small neighborhood of $x_0$, hence $f$ analytic at $x_0$. 
In particular, the set $K$ where $f$ is not analytic is obviously closed (that is always true) and with no isolated element, hence it is a perfect non-empty set, so by the Baire Category Theorem if $K=\cup K_n$ there is a closed nondegenerate interval $I$ and an $n_0$ s.t $K_{n_0}\cap I$ is dense in (the nonempty) $K \cap I$
So now we have the tool we need to locally uniformize $n(t)$ at least for $K$ as we decompose $K=\cup K_n$ where $K_n$ is the set of points $t \in K$ where $|\frac{f^{(m)}(t)}{m!}| \le R^m, m \ge n$
So we find a closed (nondegenerate) interval $I$ and $n_0$ fixed s.t $|\frac{f^{(m)}(t)}{m!}| \le R^m, m \ge n_0, t \in K_{n_0} \cap I$ and $K_{n_0} \cap I$ is dense in $K \cap I$ which by continuity means that the above inequality holds everywhere in $K \cap I$ (which is non-empty); we can ssume wlog that $I$ has length less than $1/(4R)$ so on any component $U$ of the (relatively) open set $I-K\cap I$, $\sum{\frac{f^{(n)}(t)}{n!}}(x-t)^n \to f(x)$ for $t, x \in U$ and also wlog by taking a larger $n_0$ we can assume that at the two ends $a,b$ of $I=[a,b]$ the inequality $|\frac{f^{(m)}(y)}{m!}| \le R^m, m \ge n_0, y=a$ or $y=b$ holds even if they are not in $K\cap I$ (since again by OP it holds for some $n(a), n(b)$)
But now if $\alpha<\beta$ are the ends of such an interval $U$ component of $I-K\cap I$, it is clear that $\sum{\frac{f^{(n)}(\alpha)}{n!}}(x-\alpha)^k \to f(x)$ since the Taylor series $g_x$ of $f$ at some $x \in U$ has radius of convergence bigger than $x-\alpha$ by hypothesis, so by continuity $g_x^{(k)}(\alpha)=f^{(k)}(\alpha)$ for all $k \ge 0$; the issue is of course that we do not know what happens to the left of $\alpha$ (and similarly to the right of $\beta$)
Let $h(x)$ any analytic function in $I$ s.t $\frac{h^{(m)}(y)}{m!} > R^m$ for all $y \in I, m \ge 0$ while $h$ has radius of convergence bigger than $b-a$ the length of $I$ at any point of $I$
(for example we can take $h(x)=\frac{c-a}{c-x}, b-a \le 1/(4R) < c-b <1/(3R), \frac{h^{(m)}(y)}{m!} \ge \frac{(c-a)}{(c-a)^{m+1}}>R^m$ since $c-a < 7/(12R)<R$)
Since then on $K\cap I$ we have $\frac{h^{(m)}(y)}{m!} > \frac{f^{(m)}(y)}{m!}, n \ge n_0$ and on any component $U$ of $I-K\cap I$ we have $\frac{h^{(m)}(y)}{m!} > \frac{f^{(m)}(y)}{m!}, n \ge n_0$ at the left end $\alpha$ and $h-f$ is analytic with taylor series given by $\frac{h^{(m)}(y)}{m!} > \frac{f^{(m)}(y)}{m!}$ at all points in $U$ plus $\alpha$, we immediately get $(h-f)^{(m)}(y) \ge 0, m \ge n_0, y \in U$ 
Hence $(h-f)^{(m)}(y) \ge 0, m \ge n_0, y \in I$
But now Bernstein Theorem for absolutely monotonic functions on an interval like $(h-f)^{(n_0)}$ on $I$ (for which my original argument using the remainder works), implies $(h-f)^{(n_0)}$ analytic on $I$, hence $h-f$ hence $f$ is analytic on $I$ and that is finally a contradiction with $K\cap I$ non-empty
Edit later - regarding $C^{\infty}$ functions that have all derivatives of constant sign there are various results due to Bernstein (and extended by various people); the simplest is that if $f^{(k)}(x) \ge 0, k \ge 0, x \in I$ then $f$ is analytic and more over if $a \in I$ where the right end of $I$ is $c>a$ $f$ can be extended to a holomorphic function on the complex disc $B(a,r), r=c-a$ where $r=\infty$ when $c=\infty$ 
Let $a<b, b \in I$; the remainder of the Taylor series for $x \in [a,b)$ is $R_n(x)=\frac{1}{(n-1)!}\int_a^x {f^{(n)}(t)(b-t)^{n-1}(\frac{x-t}{b-t})^{(n-1)}}dt \le (\frac{x-a}{b-a})^{(n-1)}\frac{1}{(n-1)!}\int_a^x f^{(n)}(t)(b-t)^{n-1}dt \le (\frac{x-a}{b-a})^{(n-1)}R_n(b) \le (\frac{x-a}{b-a})^{(n-1)}f(b)$ where all the inequalities follow from the non-negativity hypotesis on the derivatives and elementary considerations (eg $R_n(b)=f(b)-f'(b)(b-a)-..\le f(b)$) so $R_n(x) \to 0, x \in [a,b)$ since $|\frac{x-a}{b-a}| <1$ This means the Taylor series around $a$ converges on $[a,b)$ to $f(x)$ and by general power series stuff it means it has radius of convergence at least $b-a$ in the complex plane; letting $b \to c$ proves the claim!
