Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-\frac{1}{n}})$, everything independent. I want to find the convergence in law of: $$ Z_n = \frac{1}{n}X_n + \beta Y_n $$
With $\beta \in \mathbb{R}$.
Now, my plan is to use the characteristic function of $Z_n$.
This is what I have done: $$\phi_{Z_n} = \phi_{\frac{1}{n}X_n} *\phi_{\beta Y_n} $$
$$\phi_{Z_n} = \frac{ e^{-\frac{1}{n}} e^{i\beta t}}{1-(1-e^{-\frac{1}{n}})e^{i\beta t} } *e^{n(e^{it} - 1)}$$
I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?
Note that defining $(W_n)_{n \in \mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$ \frac{1}{n} X_n \overset{\mathcal{D}}{=} \frac{1}{n} \sum_{i=1}^n W_i $$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $\beta Y_n \to \beta$ in probability since for any $\varepsilon > 0$
$$ P(|\beta Y_n - \beta| \geq \varepsilon) = P\left(|Y-1| \geq \frac{\varepsilon}{|\beta|} \right) \leq \frac{\textrm{Var}(Y_n) \beta^2}{\varepsilon^2} $$
and since $\textrm{Var}(Y_n) = \frac{1-e^{-\frac{1}{n}}}{(e^{-\frac{1}{n}})^2 } \to 0$ as $n \to \infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.