I am trying to solve the following:
Suppose $X_{nk}, k=1,2,\ldots,n, n≥ 2$ are i.i.d. random variables $$P(X_{nk}=0)=1-\frac{1}{n}-\frac{1}{n^2}\\P(X_{nk}=1)=\frac{1}{n}\\P(X_{nk}=2)=\frac{1}{n^2}$$ Let $S_n=\sum_{k=1}^n X_{nk}$. Find the probability generating function of $X_{nk}.$ Then show that $S_n \to \mathrm{Poisson}(1)$ in distribution.
I was trying to figure out a way to recover the probability generating function by applying the fact the the probability mass function can be determined by taking successive derivatives of the generating function, but I can't seem to figure it out.
For the second step I assume once I have the probability generating function then the generating function for is $G_{S_n}=\prod_{k=1}^{n} G_{X_nk}=(G_{X_nk})^n$ since the $X_{nk}$ are iid. Then the lim as $n \to \infty$ should end up being $e^{s+1}$ which corresponds to the generating function for $\mathrm{Poisson}(1)$.
$$G_{S_n}(z)=\left[G_{X_{n1}} (z)\right]^n=\left[\mathbb{E}z^{X_{n1}}\right]^n=\left[\frac{z}{n}+\left(\frac{z}{n}\right)^2+\left(1-\frac{1}{n}-\frac{1}{n^2}\right)\right]^n$$
$$\lim_{n\rightarrow \infty}G_{S_n}(z)=e^{z-1}$$
which corresponds to the probability generating function of $\text{Poisson(1)}$.
Note: For large $n$
$$\left[1+\frac{1}{n}(z-1)+\frac{1}{n^2}(z^2-1)\right]^n\approx e^{\left[\frac{1}{n}(z-1)+\frac{1}{n^2}(z^2-1)\right]\times n}=e^{(z-1)+o(1)}$$