For $z \in \Bbb{C}$, $n\geq 1$ let, $$a_n(z) := e^{-z/n}\left(e^{z/n}-1-\dfrac{z}{n}\right)$$ Then, for $z \neq 0$, $$ \left(\dfrac{n}{z}\right)^2 a_n(z) \xrightarrow[n\rightarrow +\infty]{} \dfrac{1}{2} \quad \text{ therefore }\quad |a_n(z)| \underset{n\rightarrow +\infty}{\sim} \dfrac{1}{2} \dfrac{|z|^2}{n^2} $$ Hence the series $\sum_{n = 1}^{+\infty} a_n(z)$ converges absolutely, but given a compact subset $K$ of $\Bbb{C}$, I feel one can get more, namely, $$\sup_{z \in K} \{ |a_n(z)| \} =: \Vert a_n\Vert_{\infty}^K = \mathcal{O} \left( \dfrac{1}{n^2} \right) $$ that would give us the uniform convergence of $\sum a_n(z)$ on any compact subset. But I cannot think of any proper explanation at the moment. Any help would be gladly appreciated.
Edit: After thinking about it, I might have found a more general approach. Let $K$ be a compact subset of $\Bbb{C}$, then as $$ f(w) := \dfrac{e^{-w}}{w^2}(e^w-1-w) \xrightarrow[w\rightarrow 0]{} \dfrac{1}{2} $$ there is $r>0$ such that $$\forall w \in D(0,r), |f(w)|\leq 1$$ As $K$ is compact, it is bounded therefore there is some $M_K>0$ such that $K \subset D(0,M_K)$. Therefore, there exists $N_K \in \Bbb{N}$ such that, $$\forall n \geq N_K, \forall z \in K, \left| \dfrac{z}{n} \right| \leq r.$$ Hence, $$\forall n \geq N_K, \forall z \in K, \quad \left|f\left(\dfrac{z}{n}\right)\right| \leq 1,$$ which yields, $$\forall n \geq N_K, \forall z \in K, \quad |a_n(z)| \leq \dfrac{|z|^2}{n^2} \leq \dfrac{M_K^2}{n^2}.$$ Therefore, $$ \forall n \geq N_K, \quad \sup_{z \in K} |a_n(z)| \leq \dfrac{M_K^2}{n^2}.$$ Whence the normal convergence of $\sum a_n(z)$ on any compact subset of $\Bbb{C}$.
One can show that the function $$ f(w) = e^{-w}(e^w-1-w) $$ satisfies $$ \tag{$*$} |f(w)| \le \frac 12 |w|^2 e^{|w|} $$ for all $w \in \Bbb C$. Setting $w = z/n$ then gives the estimate $$ |a_n(z)| \le \frac 12 \frac{|z|^2}{n^2} e^{|z|/n} \le \frac 12 \frac{|z|^2}{n^2} e^{|z|} $$ and since $|z|^2e^{|z|}$ is bounded on a compact set $K$, the desired conclusion $$ \sup_{z \in K} \{ |a_n(z)| \} = \mathcal{O} \left( \dfrac{1}{n^2} \right) $$ follows.
It remains to prove $(*)$: First we determine the power series of the function $f$: $$ \begin{align} f(w) &= 1-e^{-w}(1+w) \\ & = 1 - \left( \sum_{k=0}^\infty \frac{(-1)^k}{k!}w^k\right) \cdot (1+w) \\ & = \sum_{k=2}^\infty (-1)^k\left( -\frac{1}{k!}+\frac{1}{(k-1)!}\right) w^k \\ & = \sum_{k=2}^\infty (-1)^k \frac{k-1}{k!} w^k \, . \end{align} $$ Now we can estimate the absolute value: $$ \begin{align} |f(w)| &\le \sum_{k=2}^\infty \frac{k-1}{k!} |w|^k \\ & = \frac 12 |w|^2 \sum_{k=2}^\infty \frac{2(k-1)}{k!} |w|^{k-2} \\ & = \frac 12 |w|^2 \sum_{k=0}^\infty \frac{2(k+1)}{(k+2)!} |w|^{k} \\ & \le \frac 12 |w|^2 \sum_{k=0}^\infty \frac{1}{k!} |w|^{k} \\ & = \frac 12 |w|^2 e^{|w|} \, . \end{align} $$