I needed to show that if a$ \displaystyle\sum c_nz^n$ has a convergence radius $R>0$ then the one of $\displaystyle\sum \dfrac{c_nz^n}{n!}$ is infinite and that its sum $f(z)$ follows $\forall 0<\theta<1 \;\exists M(\theta)\geq0$ such as $\lvert f(z) \rvert \leq M(\theta)e^{\frac{\lvert z\rvert}{\theta R}}$.
I did the first part all right but i can't seem to get the right inequality for the sum $f(z)$
The statement about the radius of convergence guarantees that for any $\epsilon\gt0$, $$ |c_n|\le\frac{a_\epsilon(1+\epsilon)^n}{R^n} $$ Then, $$ \begin{align} \left|\sum_{n=0}^\infty\frac{c_nz^n}{n!}\right| &\le a_\epsilon\sum_{n=0}^\infty\frac1{n!}\left(\frac{1+\epsilon}R|z|\right)^n\\ &\le a_\epsilon e^{\frac{1+\epsilon}R|z|} \end{align} $$ Set $\epsilon=\frac1\theta-1$.