Sufficient condition for a Markov chain with tridiagonal transition matrix to be null recurrent Consider the random walk induced by the Markov matrix
\begin{equation}
\begin{Vmatrix}P_{ij}\end{Vmatrix}=
\begin{Vmatrix}
r_0 & p_0 & 0 & 0 & \cdots \\
q_1 & r_1 & p_1 & 0 & \cdots \\
0 & q_2 & r_2 & p_2 & \cdots \\
\end{Vmatrix}
\end{equation}
Let
\begin{equation}
\pi_0=1,\pi_n=\frac{p_0p_1\cdots p_{n-1}}{q_1q_2\cdots q_n}.
\end{equation}
Assume that the Markov chain is recurrent. Show that
\begin{equation}
\sum^\infty_{i=0}\pi_i=\infty\Longrightarrow\mathrm{null\;reccurent}
\end{equation}
\begin{equation}
\sum^\infty_{i=0}\pi_i<\infty\Longrightarrow\mathrm{positive\;reccurent}
\end{equation}
I have completely no idea. Please give me a hint.
 A: If $\lim_{n\rightarrow\infty}P^n_{ii}=b_i>0$ for one $i$ in an aperiodic recurrent class, then $b_j>0$ for all $j$ in the class of $i$. In this case, we call the class positive recurrent or strongly ergodic. If each $b_i=0$ and the class is recurrent we speak of the class as null recurrent or weakly ergodic.
Theorem In a positive recurrent aperiodic class with states $j=0,1,2,\cdots$,
\begin{equation}
\lim_{n\rightarrow\infty}P^n_{jj}=b_j=\sum^\infty_{i=0}b_iP_{ij},\;\sum^\infty_{i=0}b_i=1
\end{equation}
and the $b$'s are uniquely determined by the set of equations
\begin{equation}
b_i\geq0,\;\sum^\infty_{i=0}b_i=1,\;\mathrm{and}\;b_j=\sum^\infty_{i=0}b_iP_{ij}
\end{equation}
We investigate the existence of a stationary probability distribution, i.e., we wish to determine the positive solutions of
\begin{equation}
(p_i+q_i)x_i=\sum^\infty_{j=0}x_jP_{ji}=p_{i-1}x_{i-1}+q_{i+1}x_{i+1},\;i=0,1,\cdots,\;(1)
\end{equation}
under the normalization
\begin{equation}
\sum^\infty_{i=0}x_i=1,
\end{equation}
where $p_{-1}=0$, and thus $x_0=q_1x_1+r_0x_0$. Using Equation (1) for $i=1$, we could determine $x_2$ in terms of $x_0$. Equation (1) for $i=2$ determines $x_3$ in terms of $x_0$, etc. It is immediately verified that
\begin{equation}
x_i=\frac{p_{i-1}p_{i-2}\cdots p_0}{q_iq_{i-1}\cdots q_1}x_0=x_0\pi_i,\;i\geq1,
\end{equation}
is a solution of (1), with $x_0$ still to be determined. Now since
\begin{equation}
1=x_0+\sum^\infty_{i=1}x_0\pi_i=\sum^\infty_{i=0}x_0\pi_i,
\end{equation}
we have
\begin{equation}
x_0=\frac{1}{\sum^\infty_{i=0}x_0\pi_i}
\end{equation}
and so
\begin{equation}
x_0>0\;\mathrm{if\;and\;only\;if}\;\sum^\infty_{i=0}x_0\pi_i<\infty.
\end{equation}
Combining this result, the definition, and the theorem, we get the desired sufficient condition.
