Consider a B&D process with infinitely many states, State 0 absorbing. Probability of ultimate absorption when starting in State $n$ (denoted $a\_n$) is computed by solving $$a\_n=\frac{\lambda\_na\_{n+1}+\mu\_na\_{n-1}}{\lambda\_n+\mu\_n}$$ for $n\ge1$, where $\lambda\_n$ and $\mu\_n$ are the up and down rates (respectively), all positive, and $a\_0=1$. Seeking a solution other than $a\_n\equiv1$ one defines $d\_n=a\_n-a\_{n+1}$ which then leads to$$d\_n=\frac{\alpha\prod\_{j=1}^n\frac{\mu\_j}{\lambda\_j}}{\sum\_{n=0}^\infty\prod\_{j=1}^n\frac{\mu\_j}{\lambda\_j}}$$ where $\alpha$ is a constant, and $$a\_n=1-\sum\_{j=0}^{n-1}d\_j$$ whenever the infinite sum converges (otherwise, $a\_n=1$ is the correct solution). To establish the value of $\alpha$, there is a 'simple probabilistic argument' (Karlin) that $\lim\_{n\to\infty}a\_n$ cannot be positive, thus must be equal to zero (implying that $\alpha=1$). My question is: what is that argument?



Computing probability of ultimate absorption in B&D processes
$\newcommand{\intr}[2]{\overline{#1,#2}}$
The desired result follows immediately from 



> 
> **Theorem** 
> 
> 
> (I) If $a\_1=1$, then $a\_j=1$ for all $j\in\intr1\infty$. 
> 
> 
> (II) If $a\_1<1$, then $S\_\infty<\infty$ and $a\_j=(S\_\infty-S\_j)(1-a\_1)$ for all $j\in\intr1\infty$, where 
>  \begin{equation\*}
>  S\_j:=\sum\_{i=1}^j s\_i,\quad s\_j:=r\_{j-1}\cdots r\_1,\quad r\_j:=q\_j/p\_j, \quad
>  p\_j:=\frac{\lambda\_j}{\lambda\_j+\mu\_j},\quad q\_j:=1-p\_j. 
> \end{equation\*}
> 
> 
> 


*Proof* Consider the embedded discrete-time Markov chain $(Y\_t)\_{t\in\intr0\infty}$, with state space $\intr0\infty$ and transition probabilities $P(Y\_{t+1}=j+1|Y\_t=j)=p\_j=1-P(Y\_{t+1}=j-1|Y\_t=j)$ for $j\in\intr1\infty$ and $P(Y\_{t+1}=0|Y\_t=0)=1$, for all $t\in\intr0\infty$. Then the probabilities of the absorption (at $0$) for the embedded chain are the same $a\_j$'s, as for the original birth-and-death process. 


The key observation is the following simple one: Fix any $j\in\intr0\infty$. For any natural $N\ge j$, let $a^N\_j$ denote the conditional probability that the embedded chain reaches the state $0$ before it reaches the state $N$ given that the chain starts in state $j$. Then, by the continuity of probability theorem ([Theorem 10.2](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=2ahUKEwiuzc6v5cDlAhUrnq0KHVksA0YQFjACegQIABAC&url=http%3A%2F%2Fstat.math.uregina.ca%2F~kozdron%2FTeaching%2FRegina%2F451Fall13%2FHandouts%2F451lecture10.pdf&usg=AOvVaw2jfikxL6LCt8zFOTpPJsYR)),  

\begin{equation}
 \text{$a^N\_j\to a\_j$.}\tag{0}
\end{equation}
Everywhere here, the convergence is as $N\to\infty$. 


Let us now compute $a^N\_j$. We have 
\begin{equation\*}
 a^N\_0=1,\quad a^N\_N=0,\quad a^N\_j=p\_ja^N\_{j+1}+q\_ja^N\_{j-1}\ \forall j\in\intr1{N-1}. 
\end{equation\*}
The latter equality can be rewritten as $h^N\_{j+1}=r\_jh^N\_j$, where 
\begin{equation\*}
 h^N\_j:=a^N\_{j-1}-a^N\_j, 
\end{equation\*}
whence 
\begin{equation\*}
 h^N\_j=r\_{j-1}\cdots r\_1h^N\_1=s\_jh^N\_1\tag{0.5}
\end{equation\*}
and, further, 
\begin{equation\*}
 a^N\_j=\sum\_{i=j+1}^Nh^N\_i=\sum\_{i=j+1}^Ns\_ih^N\_1=(S\_N-S\_j)(1-a^N\_1) \tag{1}
\end{equation\*}
for all $j\in\intr0N$. In particular, for $j=0$ formula (1) yields 
\begin{equation\*}
 1=S\_N(1-a^N\_1). 
\end{equation\*}
So, by (0), 
\begin{equation}
 a\_1=1\iff S\_\infty=\infty. \tag{2}
\end{equation}


Now consider the following two cases: 


\*Case I: $a\_1=1$. Then $a^N\_1\to1$, $h^N\_1\to0$, and hence, by (0.5), $h^N\_j\to0$ for all $j\in\intr1\infty$. So, 
$a^N\_j=1-\sum\_{i=1}^j h^N\_i\to1$ and hence $a\_j=1$ for all $j\in\intr1\infty$. This proves part (I) of the theorem. 


\*Case II: $a\_1<1$. Then part (II) of the theorem follows immediately from (2), (1), and (0). 


