This is computed based on the following recursive formula $$w\_n=\frac{\lambda\_nw\_{n+1}+\mu\_nw\_{n-1}+1}{\lambda\_n+\mu\_n}$$ where: $n$ is the inital state, State $0$ is absorbing, $\lambda\_n$ and $\mu\_n$ are the up and down rates respectively and $$\sum\_{n=0}^\infty\prod\_{j=1}^n\frac{\mu\_j}{\lambda\_j}$$diverges (to make extinction certain). To get the recursion started, we need $w\_0=0$ and $$w\_1=\frac{1}{\mu\_1}\sum\_{n=0}^\infty\prod\_{j=1}^n\frac{\lambda\_j}{\mu\_{j+1}}$$The derivation of the last formula can be found in S. Karlin's classic book "A first course in stochastic processes". The last step of his proof requires showing that $$\lim\_{n\to\infty}\prod\_{j=1}^n\frac{\lambda\_j}{\mu\_j}(w\_n-w\_{n+1)}=0$$To prove that is, according to Karlin, "more involved but still possible" (but he does not do it). How does one prove that the last limit must equal to $0$?



Expected time till extinction in a B&D process
Clearly, the expected time $w\_n$ till extinction from the initial state $n$ is nondecreasing in $n$. So, if $w\_1=\infty$, then $w\_n=\infty$ for all natural $n$, so that the difference $w\_{n+1}-w\_n$ makes no sense, and hence 
the desired conclusion 
\begin{equation}
\lim\_{n\to\infty}\prod\_{j=1}^n\frac{\lambda\_j}{\mu\_j}(w\_n-w\_{n+1)}=0 \tag{1} 
\end{equation}
makes no sense either. 


It remains to consider the case $w\_1<\infty$. Then the equation 
\begin{equation}
 w\_n=\frac{\lambda\_nw\_{n+1}+\mu\_nw\_{n-1}+1}{\lambda\_n+\mu\_n} \tag{2}
\end{equation}
(together with the condition $w\_0=0$) 
implies that $w\_n<\infty$ for all natural $n$. 
For natural $j$, let then 
$$h\_j:=w\_j-w\_{j-1},\quad\pi\_j:=\prod\_{i=1}^{j-1}\frac{\lambda\_i}{\mu\_i} 
$$
(so that $\pi\_1=1$), and 
$$u\_j:=\pi\_j h\_j.$$
Then (2) can be rewritten as 
$$u\_{n+1}=u\_n-\frac{\pi\_n}{\mu\_n}, 
$$
which implies
\begin{equation}
u\_n=u\_1-\sum\_{j=1}^{n-1}\frac{\pi\_j}{\mu\_j}. \tag{3} 
\end{equation}


Also, the equality 
$$w\_1=\frac{1}{\mu\_1}\sum\_{n=0}^\infty\prod\_{j=1}^n\frac{\lambda\_j}{\mu\_{j+1}}$$
can be rewritten as 
$$u\_1[=h\_1=w\_1]=\sum\_{j=1}^\infty\frac{\pi\_j}{\mu\_j}. 
$$
This and (3) imply
$$\pi\_n h\_n=u\_n=\sum\_{j=n}^\infty\frac{\pi\_j}{\mu\_j}. 
$$
Now the case condition $w\_1<\infty$ implies $\pi\_n h\_n\underset{n\to\infty}\longrightarrow0$, which means that the desired conclusion (1) holds. 


