Suppose a complex series with complex coefficients $$a=\sum_{n=0}^{\infty}c_{n}z^{-n-1}$$ Then its Borel transform is defined by $$Ba(x)=\sum_{n=0}^{\infty}\frac{c_{n}}{n!}x^n$$ ($x$ is a complex number).
The claim is that if the series $a$ has a non-zero radius of convergence, then the Borel transform $Ba(x)$ is an entire function(that is, holomorphic everywhere in the complex plane) of exponential type(its growth is bounded by an exponential function).
If the radius of convergence of the Borel tranform is finite, then that of the original series is zero.
How does one prove these claims?
This question is old, but an answer may help someone in the future.
Let $ f(z) $ be a holomorphic function, $ f(z) = \sum_{n=0}^\infty a_n z^n $ with a convergence radius $R < \infty$, $ Bf(z) = \sum_{n=0}^\infty \frac{a_n}{n!} z^n$.Using Cauchy's root test we see that $$ \limsup\limits_{n \rightarrow \infty}\; (a_n)^{1/n} = \frac{1}{R}, \quad\Rightarrow\quad \limsup\limits_{n \rightarrow \infty}\; (\frac{a_n}{n!})^{1/n} = 0 $$ so the radius of convergence is $\infty$.
We have by Cauchy's inequality for Taylor coefficients that for some radius $ r < R $ $$ |a_n| r^n \leq \max_{|z|=r} |f(z)| \quad\Rightarrow\quad \frac{|a_n|}{n!} \leq \frac{1}{r^n n!}\max_{|z|=r} |f(z)|. $$
Using this we get \begin{align} |Bf(z)| & = | \sum_{n=0}^\infty \frac{a_n}{n!} z^n | \\ & \leq \sum_{n=0}^\infty \frac{|a_n|}{n!} |z|^n \\ & \leq \sum_{n=0}^\infty \frac{1}{r^n n!} |z|^n \max_{|z|=r} |f(z)| \\ & = \max_{|z|=r} |f(z)| \; \sum_{n=0}^\infty \frac{1}{n!} (\frac{|z|}{r})^n \\ & = \max_{|z|=r} |f(z)| \; e^{|z|/r}. \\ \end{align}
Note that this holds true for all $z \in \mathbb{C}$ and that $r$ is a constant. Therefore, $Bf$ is bounded by an exponential function.