Problem: Use the continuity's theorem for to solve this problem: Let $X_{1},X_{2}, \overset{i.i.d}{\sim}\textbf{Poisson}(\lambda)$.Also, let $S_{n}=\sum_{i=1}^{n}X_{i}$. Prove that $$\frac{S_{n}}{n}\overset{D}{\to}\lambda$$
My attempt: I know that result is inmediate of the Weak Law of Large Numbers (WLLN), by I need to use the continuity's theorem.
Now, I know that the continuity'theorem statement that: $$F_{X_{n}}\overset{D}{\to} F_{X}(x) \iff M_{X_{n}}(t) \to M_{X}(t) \quad \text{as} \quad n\to \infty$$ where $F_{X}(x)$ is the CDF and $M_{X}(t)$ is the generating moment function.
But, \begin{eqnarray*} M_{Z_{n}}(t)=\mathbb{E}[e^{tZ_{n}}]=\mathbb{E}[e^{tS_{n}/n}]=\mathbb{E}[e^{\frac{tS_{n}}{n}}]=[\mathbb{E}[e^{tX}]]^{n}=[e^{\lambda(e^{t}-1)}]^{n} \end{eqnarray*}
But, I don't know how I would to conclude, because $\displaystyle \lim_{n \to \infty}M_{Z_{n}}(t)\not=\lambda$
What is the mystake? What am I not understanding well?
You are missing a $\frac{t}{n}$ instead of $t$ in the expectation: $M_{Z_n}(t) = \big[\mathbb{E}\big(e^{\frac{t}{n}X}\big)\big]^n = \big[e^{\lambda (e^{\frac{t}{n}}-1)}\big]^n$.
Since $n(e^{\frac{t}{n}}-1) \to t$, you get $M_{Z_n}(t) \to e^{\lambda t}$.
Since the MGF of the constant variable $\lambda$ is $t \mapsto e^{\lambda t}$, $Z_n$ converges in distribution to $\lambda$.