We know that for a Markov Chain, the condition that a state $i$ is transient implies $\lim_{n\to \infty}p_n(i,i)=0$ where $p_n$ is the n-step transition probability. Is the reverse true or not? I know there are counterexamples when the chain has an infinite state space, e.g. simple random walk on $\mathbb{Z}$.
Then on finite state space, can $p_n(i,i) \to 0$ imply that $i$ is transient?
If we have a finite state space $V$, if a state $i$ is recurrent, it is in some recurrent class $\mathcal C$, which is also a closed class.
For any $j\in\mathcal C$, this means that there exists some minimal length path $jk_{1j}\ldots k_{mj} i$ from j to i. Using this, $p_n(i,i)$ is lower bounded by $\min_{j\in\mathcal C}p(j,k_{1j})p(k_{1j},k_{2j})\cdots p(k_{mj},i)>0$ for any $n\in\mathbb N$ that is larger than $|V|$, the cardinality of your vertex set. So your state cannot be recurrent. Since any state is either recurrent or transient, your result follows.
Note that a recurrent class is positive recurrent for finite state spaces. Hence if we restrict the state space to such a class, there exists a stationary distribution for this chain. By standard results on convergence to distribution, see e.g. Norris' book, $\S1.8$, it is even possible to quantify $\lim_{n\to\infty}p_n(i,i)>0$.
