Well, let's try this... Let $\{N_{t}\}_{t\geq0}$ be an inhomogeneous Poisson process with rate function $\{\lambda(t)\}_{t\geq0}$. Let's try to compute the transition probabilites.
By definition of an inhomogeneous Poisson process, we have
$$P\{N(t+h)-N(t)=1\}=\lambda(t)h+o(h)$$
$$P\{N(t+h)-N(t)>1\}=o(h)$$
Since $\{N_{t}\}_{t\geq0}$ has independent increments by definition, we have
$$P\{N(t+h)-N(t)=1|N(t)=k\}=\lambda(t)h+o(h)$$
$$P\{N(t+h)-N(t)>1|N(t)=k\}=o(h)$$
for any $t\geq0$, any $k\in\mathbb{N}_{0}$ and "small" $h$.
Dividing previous equations by $h$, and letting $h\rightarrow 0$, we get
$$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=i+1|N(t)=i\}=\lambda(t)$$
$$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=j|N(t)=i\}=0$$
for $j>i+1$ or $j<i$.
Your statements are slightly confusing, since a transient chain could be considered "non null recurrent" (after all, it is not null recurrent). So you should replace your statements with "positive recurrent":
if all states of an irreducible Markov chain are positive recurrent, then the MC has a unique stationary distribution
if an irreducible Markov chain is finite, then all of its states are positive recurrent.
You can also say that if a Markov chain is irreducible and positive recurrent, then the (unique) stationary distribution has strictly positive components.
Every finite state irreducible Markov chain $\{M(t)\}_{t=0}^{\infty}$ has a unique stationary distribution $\pi =(\pi_i)_{i \in S}$ (where $S$ denotes the finite state space). When you simulate, with probability 1, the sample path fractions of time converge to this distribution, so that:
$$ \lim_{T\rightarrow\infty} \frac{1}{T}\sum_{t=0}^{T-1} 1\{M(t)=i\} = \pi_i \quad, \forall i \in S \quad \mbox{(with prob 1)}$$
regardless of the initial state $M(0)$. Taking expectations of both sides and using the bounded convergence theorem together with $E[1\{M(t)=i\}]=P[M(t)=i]$ we also get:
$$ \lim_{T\rightarrow\infty} \frac{1}{T}\sum_{t=0}^{T-1} P[M(t)=i] = \pi_i \quad, \forall i \in S$$
If the chain is finite state, irreducible, and also aperiodic, you can further say
$$ \lim_{n\rightarrow\infty} P[M(t)=i|M(0)=j] = \pi_i \quad, \forall i \in S$$
regardless of the initial state $j \in S$. So if the chain is finite state, irreducible, but limiting probabilities do not converge, then the chain cannot be aperiodic.
If $M(t)$ is finite state, irreducible, and periodic with period $d>1$, then limiting probabilities cannot converge (assuming we start in a particular state with probability 1). This is because:
\begin{align*}
\lim_{k\rightarrow\infty} P[M(kd)=i|M(0)=i] > 0 \\
\lim_{k\rightarrow\infty} P[M(kd+1)=i|M(0)=i] = 0
\end{align*}
This is because the Markov chain $\{Z(k)\}_{k=0}^{\infty}$ defined by $Z(k)=M(kd)$ is irreducible and aperiodic (over an appropriately reduced state space) and so all states it can reach have positive steady state values.
With this reasoning, it can be shown that $P[M(t)=i|M(0)=j]$ converges (as $t\rightarrow\infty$) to a periodic function with period $d$. The particular $d$-periodic function it converges to depends on the intial state.
This should be pretty clear if you do a few matlab examples: Plot $\vec{\pi}(t) = \vec{\pi}(0)P^t$ versus $t \in \{0, 1, 2, ...\}$ for some examples with $\vec{\pi}(0)=[1 , 0, 0, ...]$ or $\vec{\pi}(0)=[0, 1, 0, 0, ...]$.
Best Answer
OKay have I have done this. I am not going to give you the solution, but I will offer you some hints:
You attempted approach to think about $p_{ii}^{(n)}$ is correct
$p_{ii}^{(0)}=1$ - agreed?
what is $p_{ii}^n$? well I think this is the probability of Poisson process jumping $n$ steps in time interval $n$. well the number of jumps in a time interval length $n$ is distributed as ....? so this probability is .... ?
The term should be a nice form $\bigg(\dfrac{n}{e}\bigg)^n/n!$ stirling's formula tell you $n!$ is approximately what? so the term behaves like $\dfrac{1}{n^\alpha}$ for $\alpha = ? <1$ therefore the series diverges.