Bound power series where each term is divided by $n!$ Suppose the radius of convergence, $R$, of the power series $\sum_0^\infty c_nz^n$ has a positive radius of convergence. How can I show that
$$\left| \sum_{1}^\infty \frac{c_nz^n}{n!} \right| \le C \cdot e^{|z|/R}$$
for some positive constant $C$?
I am thinking we're supposed to use the complex power series expansion of $e^z = \sum_0^\infty \frac{z^k}{n!} = e^z$. I was thinking I could make a similar argument to the one made in this post Showing that a power series is bounded and multiply by something like $e^z$ instead of x (in the post). How am I supposed to use the fact that the power series $\sum_0^\infty c_nz^n$ has a positive radius of convergence?
Thank you so much!
 A: The result is not true as stated = one can only show (see comment by @Calvin or my answer using convolution to another MSE question) the result for any $0<p<1$ namely that under OP hypothesis there is $C(p)>0$ st:
$\left| \sum_{1}^\infty \frac{c_nz^n}{n!} \right| \le C(p) \cdot e^{|z|/(pR)}$
For example choose $c_0=c_1=0, c_n=n^{n/\log^2 n}, n \ge 2$; it is easy to see (taking logarithms) that $c_n^{1/n} \to 1$ so the radius of convergence of $\sum_{0}^\infty c_nz^n$ is $1$ but if we assume there is a $C>0$ for which:
$\left| \sum_{1}^\infty \frac{c_nz^n}{n!} \right| \le C \cdot e^{|z|}$ we take $z=m$ and by positivity of the coefficients we have:
$\frac{c_mm^m}{m!} < \left| \sum_{1}^\infty \frac{c_nm^n}{n!} \right| \le C \cdot e^{m}$
This gives: $m^{m/\log^2 m}m^m<Cm!e^m \le Cem^m\sqrt m e^{-m}e^m$ by an easy inequality that follows from Stirling approximatiom, which reduces to:
$m^{m/\log^2 m} < Ce \sqrt m$  and that is not possible for arbitrary large $m$ since $m/\log^2m \to \infty$ hence $m^{m/\log^2 m}/\sqrt m \to \infty$.
(It is instructive to see how this counterexample fails when we have $e^{m/p}, 0<p<1$ on RHS rather than $e^m$)
