I am working on the following problem:
Suppose $X_1, X_2,\dots$ are i.i.d Poisson$(1)$ random variables, and let $S_n=X_1+\dots+X_n$.
a)Compute $E\left[\left( \frac{S_n-n}{\sqrt{n}} \right)^-\right]$ exactly, where $x^- = (-x)\lor 0.$
b) Explain why $\left( \frac{S_n-n}{\sqrt{n}} \right)^- \implies N^-,$ where $\implies$ denotes convergence in distribution, and $N\sim N(0,1).$
c) Show that $E\left[\left( \frac{S_n-n}{\sqrt{n}} \right)^-\right]\rightarrow E(N^-)=\frac{1}{\sqrt{2\pi}}$.
d) Conclude that $n! \sim n^n e^{-n}\sqrt{2\pi n}$
So far for part (a) I was able to compute that $E\left[\left( \frac{S_n-n}{\sqrt{n}} \right)^-\right]=\frac{e^{-n}n^{n+1/2}}{n!}$.
For part (b) since $x^-$ is a continuous function and $\frac{S_n-n}{\sqrt{n}}\implies N\sim N(0,1)$ we get that $\left( \frac{S_n-n}{\sqrt{n}} \right)^-\implies N^-$.
Once I prove (c), (d) follows through manipulation of the two expectations.
I am stuck on proving (c) and could use some help. By (b) I know that $\left( \frac{S_n-n}{\sqrt{n}} \right)^-\implies N^-$, if I can show that $\{\left( \frac{S_n-n}{\sqrt{n}} \right)^-\}_n$ is uniformly integrable then I get $E\left[\left( \frac{S_n-n}{\sqrt{n}} \right)^-\right]\rightarrow E(N^-)$.
I could use help showing that $\{\left( \frac{S_n-n}{\sqrt{n}} \right)^-\}_n$ is uniformly integrable.
Thanks!
UPDATE: So my professor told me the following to help with this problem. If $\{\left|\frac{S_n-n}{n}\right|\}_n$ is uniformly integrable then $\{\left( \frac{S_n-n}{\sqrt{n}} \right)^-\}_n$ is uniformly integrable. Which is true because $\left( \frac{S_n-n}{\sqrt{n}} \right)^-\leq \left|\frac{S_n-n}{n}\right|$.
Then I was told that I should recall that $S_n$ can be written as the sum of $n$ i.i.d random variables. So now I am stuck trying to show that $\{\left|\frac{S_n-n}{n}\right|\}$ is uniformly integrable.
$\left( \frac{S_n-n}{\sqrt{n}} \right)_{n \geq 0}$ is uniformly integrable because it is bounded in $\mathbf{L}^2$.
$\mathbf{E}\bigg[\left( \frac{S_n-n}{\sqrt{n}} \right)^2\bigg] = \frac{1}{n}\mathbf{E}\big[\left( S_n-n \right)^2\big] = \frac{1}{n}\mathbf{Var}[S_n] = \mathbf{Var}[X_1]$ because the $(X_n)_{n \geq 0}$ are i.i.d., so $\sup_\limits{n \geq 1}\mathbf{E}\bigg[\left( \frac{S_n-n}{\sqrt{n}} \right)^2\bigg] < \infty$.