Let $\left(\varepsilon_{n}\right)_{n \in \mathbb{N}}$ be a sequence of idependent random variables with $0$ mean and finite variance, say $\sigma^{2}$, and let $\alpha$ be a real number such that $0 < \left| \alpha \right| < 1$. Now, for every $n \in \mathbb{N}$, set $$X_{n} = \sum_{j=1}^{n}{ \alpha^{n-j} \varepsilon_{j}}.$$ When the $\varepsilon_{n}$'s are normally distributed, it can be shown that the characteristic function of $X_{n}$ is $$\varphi_{X_{n}}{\left( t \right) } = \exp{\left(-\frac{1}{2} \frac{\sigma^{2} \left(1-\alpha^{2n} \right) }{1-\alpha^{2} } t^{2} \right)} $$ and, thus, we obtain $$ \varphi_{X_{n}} \left(t\right) \to \exp {\left(-\frac{1}{2} \frac{\sigma^{2}}{1-\alpha^{2}}t^{2} \right) } \text{ as } n\to+\infty. $$ My question is: Does the limit above still hold if we just merely assume that the $\varepsilon_{n}$'s have $0$ mean and the same variance, say $\sigma^{2}$? So far, I think the answer is negative, but I could not find an example for it.
Any help would be appreciated. Thank you in advance.
If you are looking for an example, take $\varepsilon_{n}$ having the values $\pm1$ each with probability $\frac12$, so $\sigma^2=1$. Also let $\alpha=\frac12$.
You will find $X_n$ converges in distribution to a continuous uniform random variable on the interval $(-2,2)$, which has a characteristic function of $\dfrac{\sin (2t)}{2t}$
This is not the same as the $\exp {\left(-\frac{2}{3} t^{2} \right) }$ your expression would give for $\sigma^2=1$, $\alpha=\frac12$