I have a Markov chain with a single absorbing state $s_{-1}$. The transient states have absorption probabilities $p_{i,-1} = 1-f_i$ and transition probabilities to the next state $p_{i,i+1} = f_i$. We also have that the starting state is $s_0$. I want to find the expected absorbing time. I've calculated something that looks somewhat reasonable, but I suspect that even if I didn't make a mistake somewhere, there is some easier way of doing this. The problem is of course that it has infinitely many states and the probability changes with the states.
Let $s_i = 1$ indicate we are at the absorbing state at iteration $i$. Then the time to absorption can be written as
$$E[\tau] = \sum_{i=1}^\infty i P(s_i=1,s_{i-1}=0) = \sum_{i=1}^\infty i P(s_i = 1|s_{i-1} = 0)P(s_{i-1}=0) = \sum_{i=1}^\infty i(1-f_{i-1})P(s_{i-1}=0),$$
but $P(s_{i-1} = 0) = P(s_{i-2} = 0)f_{i-2} = P(s_{i-3}=0)f_{i-2}f_{i-3} = \dots = P(s_1 = 0)\prod_{n=1}^{i-1} f_{i-n}$, and $P(s_i = 0) = 1$, so $P(s_{i-1} = 0) = \prod_{n=1}^{i-1}f_{i-n}$. Plugging in above we get
$$E[\tau] = \sum_{i=1}^\infty i (1-f_{i-1})\prod_{n=1}^{i-1}f_{i-n}.$$
This looks rather ugly of course. Can it be done more elegantly?
(PS. In my special case, we have $f_i = \binom{k+i+1}{k-1}^{-1}$ which of course goes to zero quite fast, causing the absorption probability $1-f_i$ to go to 1 quite fast as well.)