I am studying "A First Course in Stochastic Process" by Samuel Karlin.
In this book, theorem 7.1 states that
Consider a birth and death process with birth and death parameters $\lambda_n$ and $\mu_n$, where $\lambda_0 = 0$ so that $0$ is an absorbing state. The mean time to absorption is
$$ \begin{cases} \infty & \text{if} & \sum_{i=1}^\infty \rho_i = \infty \\ \sum_{i=1}^\infty \rho_i + \sum_{r=1}^{m-1} \Bigr( \prod_{k=1}^r\frac{\mu_k}{\lambda_k} \Bigl) \sum_{j=r+1}^\infty \rho_j & \text{if} & \sum_{i=1}^\infty \rho_i < \infty \\ \end{cases} $$ where $\rho_i = \lambda_1\lambda_2.... \lambda_{i-1}/\mu_1\mu_2 .... \mu_i$
In the proof of the theorem, the book states that bellow statement.
If $\sum_{i=1}^\infty \rho_i < \infty$, then $$ \lim\limits_{m \to \infty} \Bigr( \prod_{j=1}^m\frac{\lambda_j} {\mu_j}\Bigl) (\omega_m - \omega_{m+1}) = 0 $$
(Here $\omega_i$ is mean absortion time starting from state i.)
The book just says "It is more involved but still possible to prove" above statement. It does not give any detailed proof or hint for above statement.
Could someone give me proof for above statement?
Bellow is the detaield proof of theorem 7.1 in the book.
Let $\omega_i$ is mean absortion time starting from state i.
The mean time the state stays in state $i$ without any change of the state is $\frac{1}{\lambda_i + \mu_i}$.
When the state changes from $i$, it moves to state $i-1$ with probability $\frac{\mu_i}{\lambda_i + \mu_i}$ or moves to state $i + 1$ with probability $\frac{\lambda_i}{\lambda_i + \mu_i}$.
Hence,
$$
\omega_i = \frac{1}{\lambda_i + \mu_i}
+ \frac{\lambda_i}{\lambda_i + \mu_i}\omega_{i+1}
+ \frac{\mu_i}{\lambda_i + \mu_i}\omega_{i-1}
$$
Where $\omega_0 = 0$
Letting $z_i = \omega_i - \omega_{i+1}$, then
$$z_i = \frac{1}{\lambda_i} + \frac{\mu_i}{\lambda_i}z_{i-1}, \qquad i \ge 1$$
$$z_m = \frac{1}{\lambda_m} + \frac{\mu_m}{\lambda_m}\frac{1}{\lambda_{m-1}} + \frac{\mu_m\mu_{m-1}}{\lambda_m\lambda_{m-1}}z_{m-2} $$
$$ z_m = \sum_{i=1}^m\Bigr(\frac{1}{\lambda_i}\prod_{j=i+1}^m\frac{\mu_j}{\lambda_j}\Bigl) + \Bigr(\prod_{j=1}^m\frac{\mu_j}{\lambda_j}\Bigl)z_0 $$ then $$ \omega_m - \omega_{m+1} = \sum_{i=1}^m\Bigr(\frac{1}{\lambda_i}\prod_{j=i+1}^m\frac{\mu_j}{\lambda_j}\Bigl) - \omega_1\Bigr(\prod_{j=1}^m\frac{\mu_j}{\lambda_j}\Bigl) \qquad (1) $$ Let $$ \rho_i = \frac{\lambda_1\lambda_2.... \lambda_{i-1}}{\mu_1\mu_2 .... \mu_i} $$ then $$ \sum_{i=1}^m\Bigr(\frac{1}{\lambda_i}\prod_{j=i+1}^m\frac{\mu_j}{\lambda_j}\Bigl) = \Bigr(\prod_{j=1}^{m}\frac{\mu_j}{\lambda_j}\Bigl)\sum_{i=1}^m\rho_i \qquad (2) $$ In terms of (2), the relation (1) becomes
$$ \Bigr(\prod_{j=1}^m\frac{\lambda_j}{\mu_j}\Bigl)(\omega_m - \omega_{m+1}) = \sum_{i=1}^m \rho_i - \omega_1 \qquad (3) $$ It is probabilistically evident that $\omega_m < \omega_{m+1}$. Hence, LHS of equation (3) cannot be positive.
If $\sum_{i=1}^\infty \rho_i = \infty$, then $\omega_1$ also must be $\infty$ in order to make RHS of equation (3) not positive.
Then, $\omega_m$ is infinite for every $m\ge1$.
Now suppose $\sum_{i=1}^\infty \rho_i < \infty$, then $$ \omega_1 = \sum_{i=1}^m \rho_i - \Bigr(\prod_{j=1}^m\frac{\lambda_j}{\mu_j}\Bigl)(\omega_m - \omega_{m+1}) $$ $$ \omega_1 = \sum_{i=1}^\infty \rho_i - \lim\limits_{m \to \infty} \Bigr(\prod_{j=1}^m\frac{\lambda_j}{\mu_j}\Bigl)(\omega_m - \omega_{m+1}) $$ It is more involved but still possible to prove that
(This is the part I can't get. How to prove this?)
$$
\lim\limits_{m \to \infty} \Bigr( \prod_{j=1}^m\frac{\lambda_j} {\mu_j}\Bigl) (\omega_m - \omega_{m+1}) = 0
$$
and then indeed
$$
\omega_1 = \sum_{i=1}^{\infty} \rho_i
$$
Equation (3) becomes
$$
\Bigr(\prod_{j=1}^m\frac{\lambda_j}{\mu_j}\Bigl)(\omega_m - \omega_{m+1}) = \sum_{i=1}^m \rho_i - \sum_{i=1}^{\infty} \rho_i
$$
$$
\omega_m - \omega_{m+1} = \Bigr(\prod_{j=1}^m\frac{\mu_j}{\lambda_j}\Bigl) (\sum_{i=1}^m \rho_i - \sum_{i=1}^{\infty} \rho_i)
$$
$$
\omega_m - \omega_{m+1} = - \Bigr(\prod_{j=1}^m\frac{\mu_j}
{\lambda_j}\Bigl) \sum_{i=m+1}^{\infty} \rho_i
\qquad (4)
$$
Suming equation (4) from $m=1$ to $m=k-1$
$$
\omega_1 - \omega_k = - \sum_{m=1}^{k-1}
\Bigr(\prod_{j=1}^m\frac{\mu_j}
{\lambda_j}\Bigl) \sum_{i=m+1}^{\infty} \rho_i
$$
$$
\sum_{i=1}^{\infty} \rho_i - \omega_k = - \sum_{m=1}^{k-1}
\Bigr(\prod_{j=1}^m\frac{\mu_j}
{\lambda_j}\Bigl) \sum_{i=m+1}^{\infty} \rho_i
$$
$$
\omega_k =\sum_{i=1}^{\infty} \rho_i + \sum_{m=1}^{k-1}
\Bigr(\prod_{j=1}^m\frac{\mu_j}
{\lambda_j}\Bigl) \sum_{i=m+1}^{\infty} \rho_i
$$
and the theorem holds.
(Of course, I have added more details than the original proof in the book)
I cannot prove how bellow equation is equal to $0$. $$ \lim\limits_{m \to \infty} \Bigr( \prod_{j=1}^m\frac{\lambda_j}{\mu_j}\Bigl) (\omega_m - \omega_{m+1}) $$ All I can prove is that the equation is upper bounded by $0$, and increases as m -> $\infty$ if $\sum_{i=1}^\infty \rho_i < \infty$.
Hence, the limit will be equal to some $ c \le 0$, but I cannot find a way to prove that $c = 0$.
I am kinda disappointed in this book. Skipping some important part in the proof by just saying some incomprehensible expression "more involved but still possible"
