Let $\alpha \geq 1$ and $X_n$ be independent random variables such that $P(X_n=2)=P(X_n=-2)=\frac 1{2n^\alpha}$ and $P(X_n=0)=1-\frac 1{n^\alpha}$. Let $S_n=\sum_{k=1}^n X_k$.
Depending on the value of $\alpha$, what are the properties of $S_n$ (convergence, asymptotic behaviour)?
$S_n$ is clearly a Markov chain on the even integers. Note that it is not time-homogeneous or stationary.
Since $E(X_n)=0$ and $V(X_n)=\frac 4{n^\alpha}$, we have $V(S_n)=4\sum_{k=1}^n \frac{1}{k^\alpha}$.
Whether $\alpha=1$ or $\alpha >1$, we have respectively $V(S_n)=O(\log n)$ and $V(S_n)=O(1)$. In both cases, by Markov's bound, for any $\epsilon >0$ and $\delta >0$, $$P(\frac{|S_n|}{ n^{1/2+\epsilon}}\geq\delta) = P(|S_n|\geq n^{1/2+\epsilon}\delta)=O\left( \frac{\log n}{n^{1+2\epsilon}}\right)$$ and since $\sum_n \frac{\log n}{n^{1+2\epsilon}} < \infty$, we get $S_n = o\left(n^{1/2+\epsilon} \right)$ a.s.
As noticed by Olivier in the comments, if $\alpha>1$, $\sum_n P(X_n\neq 0) = \sum_n \frac{1}{n^\alpha}<\infty$ thus by Borel-Cantelli lemma, $P(\limsup_n (X_n\neq 0)) = 0$, i.e. $$P(\liminf_n (X_n= 0)) = 1$$ Hence almost surely, $S_n$ becomes constant.
Olivier also noticed that $S_n$ is a martingale, and for $\alpha>1$, $E(S_n^2) = V(S_n)=O(1)$. A result from the theory of martingales implies that $S_n$ converges almost surely and also converges in $L^2$.
The characteristic function of $S_n$ is $$\prod_{k=1}^n \frac 1{2k^\alpha} e^{2it} + \frac 1{2k^\alpha} e^{-2it} + 1-\frac 1{k^\alpha} = \prod_{k=1}^n \left(1-\frac{1-\cos(2t)}{k^\alpha}\right)$$ When $\alpha=1$, this converges pointwise to $$t\mapsto 1_{\pi \mathbb Z}(t) $$ This limit is not a continuous function, so Lévy's continuity theorem does not apply. This rather indicates that $S_n$ does not converge in distribution when $\alpha=1$.