sequences and series - Maximal open set where $(e^{nz}/n)_{n}$ converges uniformly

I would like to find the maximal open set such that the sequence $\left(\frac{e^{nz}}{n}\right)_{n\in\Bbb{N^*}}$ converge uniformly for $z\in\Bbb{C}$.

We have $\exp(nz)=\sum_{k=0}^{\infty}\frac{n^kz^k}{k!},$ so that $\frac{e^{nz}}{n}=\sum_{k=0}^\infty\frac{n^{k-1}z^k}{k!}.$

If I denote $a_k=\frac{n^{k-1}}{k!}$ we have $a_k\ne 0$ for all $k\in\Bbb{N.}$ We have

$$\frac{a_{k+1}}{a_k}=\frac{n^{k}k!}{n^{k-1}(k+1)!}=\frac{n}{k+1}\to0 \quad\mbox{as}\quad k\to\infty.$$

Then the radius of convergence of $\sum_{k=0}^\infty\frac{n^{k-1}z^k}{k!}$ is $\infty,$ so that it converges everywhere.

So if I am not mistaken the sequence converge uniformly on $\Bbb{C}.$

Am-I correct ?

Answer

Let $f_n(z)=\frac{e^{nz}}{n}$. Let $z=x+iy$, where $x=\text{Re}(z)$ and $y=\text{Im}(z)$. Then, we can write

$$\frac{e^{nz}}{n}=\frac{e^{nx}\,e^{iny}}{n} \tag 1$$

Clearly, for $x>0$, the sequence $f_n(z)$ diverges.

If $x\le 0$, then we have for any given $\epsilon>0$

$$\begin{align} \left|\frac{e^{nz}}{n}\right|&=\frac{e^{-n|x|}}{n}\\\\ &\le \frac1n\\\\ &<\epsilon \end{align}$$

whenever $n>N>\frac{1}{\epsilon}$. Hence the sequence converges uniformly for $x\le 0$ and diverges otherwise.

The maximum open set is therefore the left-half plane.