Exponential convergence of Taylor series. Let $f:A\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be an analytic function on the open set $A$: for each $x_0\in A$, we can find an interval $I_{x_0,\delta}=[x_0-\delta,x_0+\delta]\subseteq A$ such that $f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ for all $x\in I_{x_0,\delta}$. My question asks about conditions on $f$ under which 
$$ \limsup_{N\rightarrow\infty}\left|f(x)-\sum_{n=0}^N a_n(x-x_0)^n\right|e^N<\infty, $$
for each $x\in I_{x_0,\delta}$. That is, there is exponential convergence of the Taylor series.
 A: Let's take $x_0=0$ for convenience. Let $R$ be the radius of convergence of $\sum_{n=0}^{\infty}a_nx^n.$ Claim: If $R=\infty,$ then
$$\tag 1 e^N|\sum_{N+1}^{\infty}a_nx^n| \to 0\, \text { for all } x.$$
Thus we have your result everywhere for familiar friends like $e^x, \cos x, \sin x,\dots$
To prove  $(1),$ fix $x\in \mathbb R.$ Choose $\epsilon>0$ such that $ |\epsilon x| < 1/(2e).$ Because the radius of convergence is $\infty,$ we have $|a_n|^{1/n} \to 0.$ Thus for $n$ large enough, $|a_n|^{1/n} <\epsilon,$ which implies $|a_n|\le \epsilon^n.$ Thus for $N$ large enough,
$$e^N|\sum_{N+1}^{\infty}a_nx^n| \le e^N\sum_{N+1}^{\infty}|\epsilon x|^n = e^N\frac{|\epsilon x|^{N+1}}{1-|\epsilon x|}$$ $$ < |x|\left (\frac{e}{2e} \right )^N\frac{1}{1-|\epsilon x|} = \frac{|x|}{1-|\epsilon x|}\left (\frac{1}{2} \right )^N\to 0.$$
This proves $(1).$ You can tell that there is room to spare in the above. I'm pretty sure $e^N$ could be replaced with $A^N$ for any large $A>0.$
Now suppose $R<\infty.$ I believe in this case you'll get exponential convergence for $|x|<r$ if $r$ is small enough, but such convergence will fail to hold in the entire interval $(-R,R).$ For example, suppose we look at the series $\sum_{n=0}^{\infty}x^n,$ which has ROC $1.$ In this case we have
$$e^N|\sum_{n=N+1}^{\infty}x^n| = e^N|x|^{N+1}\frac{1}{1-x}.$$
That is bounded in $N$ iff $e|x|\le 1$ or $|x|\le 1/e.$
