Let $p_0=1, 0<p_n<1$ a sequence of probabilities.
define: $[A_{i,j}]_{i,j\in\mathbb N_0}$, $A_{0,0}=1$ , $A_{i,i-1}=1-p_i$ ,$A_{i,i+1}=p_i$
definition: $h:\mathbb N\rightarrow \mathbb R$ will be called harmonic at $n$ if $h_n=\sum_{k\in\mathbb N_0}[(A_{n,k})\cdot h_k]$.
I have already proven that $h$, harmonic on all $\mathbb N$ [without $0$] exists, and unique if $h_0,h_1$ are given. and dependant only on $p_1;\dots;p_{n-1}$
Now I need to prove that the chain is recurrent iff $\lim_{n\rightarrow\infty}h_n=\infty $.
Definition: a markov chain will be called recurrent at $m$ if starting at $X_n=m$: $P(min\{k\geq 1,X_{n+k}=m\})=1$ [the probability of returning to $m$ after arriving at $m$ in finite time is $1$.
I think I should be using Doob's Optional Stopping Time Theorem, but I have no idea how. Please Help.
Let $h$ be the (non-negative) harmonic function with $h(0)=0$ and $h(1)=1$. Start the Markov chain in state $1$, and consider the martingale $h(X_n)$. Let $T_k$ be the first hitting time of state $k$. Then by optional stopping $$ 1=E[h(X_0)]\ge E[h(X_{T_0\wedge T_N})]=h(N)\cdot P[T_N<T_0], $$ where $N\in\Bbb N$ is large. From this it is clear that if $\lim_Nh(N)=\infty$ then $P[T_0=\infty]=\lim_NP[T_N<T_0]=0$, and the chain is recurrent. Conversely, if $\lim_Nh(N)<\infty$ then $h$ is bounded, as is the martingale $h(X_n)$, so the inequality in the above display is an equality. Letting $N\to\infty$ you see that $P[T_0=\infty]=\lim_NP[T_N<T_0]>0$, and the chain cannot be recurrent.