I am trying to solve the following exercise: Let $X_1^{(n)},\ldots, X_n^{(n)}$ be iid r.v. such that $P(X_i^{(n)}=1)=P(X_i^{(n)}=-1)=\frac{1}{2n}, (P(X_i^{(n)}=0)=1-\frac1n$
Let $S_n = \sum_{i=1}^n X_i^{(n)}$, show that $S_n$ converges in Distribution and compute its limit.
I am trying using characteristic functions. I have that $$ E(e^{itX_1^{(n)}})= \frac{e^{it}}{2n}+\frac{e^{-it}}{2n}+1-\frac1n$$ and $$ E(e^{itS_n}) = \left(E\left(e^{itX_1^{(n)}}\right)\right)^n = \bigg(1 + \frac{e^{it}+e^{-it}-2}{2n}\bigg)^n \to \exp\bigg(\frac{e^{it}+e^{-it}-2}{2}\bigg)$$ as $n$ approaches $+\infty$. Even re-writing the limiting function as $\exp(\cos(t)-1)$ I have no Idea on how to compute the limiting random variable. How do I compute the corresponding integral? $$ \int_\mathbb{R} e^{-itx+\cos(t)-1} dt$$
Did I do something wrong?
I see nothing wrong with the computations.
In terms of known probability distributions, the limit is the difference of two independent Poisson distributed random variables with parameter $1/2$.