I'm trying to solving the following exercise and I need a check on my reasoning
Let $X \geq0$ a.s. and $t>0$.
Then $$\lim_{a \rightarrow \infty} \sum_{n=1}^{at} \frac{a^n}{n!} E[X^n e^{-aX}] = E\left[\lim_{a \rightarrow \infty}\sum_{n=1}^{at} \frac{(aX)^n}{n!} e^{-aX}\right].$$
Of course what I need to do is just to take the limit inside the expectation.
First, I notice that the limit is w.r.t $a$. Since I have a sum of positive terms, I want to apply the monotone convergence theorem.
First, before taking the limit, I bring the sum inside the expectation. $$\lim_{a \rightarrow \infty} \sum_{n=1}^{at} \frac{a^n}{n!} E[X^n e^{-aX}] = \lim_{a \rightarrow \infty} E[\sum_{n=1}^{at} \frac{a^n}{n!} E[X^n e^{-aX}]$$
My question is: to which sequence of functions do I apply the monotone convergence theorem?
I would say to the following one:
$$g(a)=\sum_{n=1}^{at} \frac{(aX)^n}{n!}e^{-aX}$$
(I just considered every term depending on $a$).
Since $g(a)$ is the partial sum of positive terms, it's obviously increasing (and obviously positive).
So I can take the limit inside the expectation by the monotone convergence theorem
Is this the way or am I missing something?