We know that irreducible Markov chains can be separated into the two cases: (1) All limits of transition probabilities vanish: $lim_{n\rightarrow\infty} p_{ij}^{(n)} = 0$ for all i,j in state space S. (2) All limits of transition probabilities do not vanish: $lim_{n\rightarrow\infty} p_{ij}^{(n)} \neq 0$ for all i,j in state space S, including cases where limits of transition probabilities do not exists.
If the chain has a stationary distribution, then it is in case (2), for example, see Vanishing Together Corollary in Rosenthal's "A First Look at Stochastic Process". Is it possible that a chain is in case (2) but has no stationary distribution?
Edit: Based on John's answer and user8675309's comment, an irreducible Markov chain has the following cases:
(1) $lim_{n\rightarrow\infty} p_{ij}^{(n)}$ exists for all i,j in S, where:
$\quad$ (1.1) $lim_{n\rightarrow\infty} p_{ij}^{(n)} = 0$ for all i,j in S, then there is no stationary distribution.
$\quad$ (1.2) $lim_{n\rightarrow\infty} p_{ij}^{(n)} > 0$ for all i,j in S, then there a stationary distribution is given by $$\pi_j = \frac{lim_{n\rightarrow\infty} p_{ij}^{(n)}}{\sum_{k \in S}lim_{n\rightarrow\infty} p_{ik}^{(n)}}$$ for all j in S.
(2) For some i,j in S, $lim_{n\rightarrow\infty} p_{ij}^{(n)}$ does not exists.
$\quad$ (2.1) There is a stationary distribution: For example, S = {1,2} and $p_{12} = p_{21} = 1$ and the stationary distribution is given by $\pi_1 = \pi_2 = \frac{1}{2}$.
$\quad$ (2.2) There is no stationary distribution.
Is (2.2) possible (assuming the chain is irreducible)? If so, what's an example?
If the chain is finite and irreducible, then all states are positive recurrent and the chain has a stationary distribution. So for (2.2) to hold, it has to be a chain with infinite state space and mean recurrence time zero for all states. I wonder whether this implies $lim_{n\rightarrow\infty} p_{ij}^{(n)} = 0$ for all i,j in S.
I know $1/m_i = lim_{n\rightarrow \infty}E_i(\frac{1}{n}\sum_{k=1}^{n}1_{X_k=i}) = lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}E_i(1_{X_k=i})=lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}P_i(1_{X_k=i})= lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}p_{ii}^{k} = 0$.
But I don't see how this imply $lim_{n\rightarrow \infty}p_{ii}^{n}=0$.
No, in case (2) the limit probabilities $\pi_j:=\lim_n p_{i,j}^{(n)}$ are all $>0$, add to $1$, and form a stationary distribution.