For a positive real $t$, what is the limit $$ \lim_{n\to\infty} a(n,t)=\lim_{t\to \infty}e^{-nt}\sum_{j=n+1}^\infty\frac{(nt)^j}{j!}. \quad\quad\quad (\star) $$ I expect it to be 0 as I should be the approximation error of the semi-discrete backward Euler method at $x=1$ of the advection equation $\partial_t u=-\partial_x u$, $t>0$, $u(0,x)=\mathbf 1_{\{x>0\}}$, i.e. $$ \partial_t u_n= -D_x^nu_n:=-n[u_n(t, x)-u_n(t, x-1/n)],\quad t>0,\quad u(0,x)=\mathbf 1_{\{x>0\}}, $$ so that
$$ u_n(t,x)-u(t,x) = e^{-tn}\sum_{j=0}^{\lfloor xn \rfloor}\frac{(tn)^j}{j!}\mathbf 1_{\{x-j/n>0\}} - \mathbf 1_{\{x-t>0\}}. $$ If $(\star)=0$, is the order of convergence $O(1/n)$?
EDIT: Due to the suggestion, one can use the following proof for convergence for fixed $0<t<1$. Using increasingness of $e^{-nz}z^n$ on $[0,n]$ \begin{align} a(n,t)=\frac1{n!}\int_0^{nt}e^{-z}z^n\,dz= \frac{n^{n+1}}{n!}\int_0^{t}e^{-nz}z^n\,dz\le t \frac{n^{n+1}}{n!} e^{-nt}t^n= t \ell_n \to 0 \end{align} as $n\to \infty$, because \begin{align} \frac{\ell_{n+1}}{\ell_n}= e^{-t}t\left(\frac{n}{n-1}\right)^n=e^{-t}t\left(1+\frac{1}{n-1}\right)^n \to e^{1-t}t<1. \end{align} Moreover, as $\sum_{n=0}^\infty a(n,t)\le C \sum_{n=0}^\infty \theta(t)^n<\infty$ for some $C>0$ and $\theta(t)$ close to $e^{1-t}t^2$, then for all large $n$ $$ a(n,t)\le C_t\frac1n. $$
I do not know how much this could help you. $$e^{-nt}\sum_{j=n+1}^\infty\frac{(nt)^j}{j!}=1-\frac{\Gamma (n+1,n t)}{ \Gamma (n+1)}$$