Show $\lim_{m \to \infty} \sup_{n \ge 0} E \left[ e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]=0$ for a sequence of positive r.v. Let $\{ X_n \}_{n \ge 0} $ be some  sequence  of non-negative radon variables.    How to show the following limit
\begin{align}
\lim_{m \to \infty} \sup_{n \ge 0} E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]=0. 
\end{align}
Clearly, we have that $e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \le e^{-X_n}  e^{X_n} =1$, so if we didn't have a $\sup$ in the expression, then the result would follow by the dominated convergence theorem. 
So,  it seems that we have to argue and get rid of the supremum.  This is the part of the problem that I am not sure how to do.
 A: In general, it is not true that 
$$
\begin{align}\tag{*}
\lim_{m \to \infty} \sup_{n \ge 0} \mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]=0. 
\end{align}
$$
For example, when $X_n$ is constant equal to $x_n$ and $x_n\to +\infty$, then 
$$
e^{-x_n}\sum_{k\gt m}\frac{x_n^k}{k!}=1-e^{-x_n}\sum_{k=1}^m\frac{x_n^k}{k!}
$$
hence for all fixed $m$, 
$$
\sup_{n\geqslant 1}e^{-x_n}\sum_{k\gt m}\frac{x_n^k}{k!}=1.
$$
In fact, a necessary and sufficient condition for (*) to hold is 
$$
\lim_{R\to \infty}\sup_{n\geqslant 1}\Pr\left\{X_n\gt R\right\}=0,\tag{**}
$$
which is usually referred as tightness.
Indeed, if $(**)$ holds, then for all fixed $R$, 
$$
\mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]=\mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!}\mathbf 1\left\{X_n\leqslant  R\right\} \right]+\mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!}\mathbf 1\left\{X_n\gt  R\right\} \right].
$$
The first term of the right hand side is bounded by $$\sup_{0\leqslant t\leqslant R}e^{-t}\sum_{k\gt m}\frac{t^k}{k!}\leqslant\sum_{k\gt m}\frac{R^k}{k!} $$
and for the second term, using the fact that $e^{-t}\sum_{k\gt m}\frac{t^k}{k!}\leqslant 1$, we get
$$
\sup_{n\geqslant 1}\mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]
\leqslant\sum_{k\gt m}\frac{R^k}{k!}+\sup_{n\geqslant 1}\Pr\left\{X_n\gt R\right\}.
$$
For the opposite direction, we use non-decreasingness of the function $t\mapsto e^{-t}\sum_{k\gt m}t^k/k!$, we get 
$$
\sup_{n \ge 0} \mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right]
\geqslant \sup_{n \ge 0} \mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!}\mathbf 1\left\{X_n\gt  R\right\}  \right]\geqslant e^{-R} \sum_{k>m} \frac{R^k}{k!} \sup_{n \ge 0} \Pr\left\{X_n\gt R\right\}
$$
hence 
$$
\sup_{n \ge 0} \Pr\left\{X_n\gt R\right\}\leqslant \left(e^{-R} \sum_{k>m} \frac{R^k}{k!} \right)^{-1}\sup_{n \ge 0} \mathbb E  \left[  e^{-X_n} \sum_{k>m} \frac{X_n^k}{k!} \right].
$$
Taking the $\limsup$ as $R$ is going to infinity and using the fact that forall fixed $m$, $e^{-R} \sum_{k>m} \frac{R^k}{k!} \to 1$, we establish (**).
