Conditions for birth and death process having only finitely many deaths. Consider a birth and death process on $\mathbb{N}=\left\{0,1,2,\ldots\right\}$, given by the transition probabilities $p(n,n+1)=\lambda_n$ and $p(n,n-1)=\mu_n$ (those are the birth and death rates, respectively), which satisfy $\lambda_n+\mu_n=1$. We assume $\lambda_n$ and $\mu_n$ to be strictly positive (so this process is irreducible).
I'm looking for conditions on the coefficients $\lambda_n$ (and $\mu_n$) that imply that, with probability $1$, almost every sample path in this system will have only births after a certain time.
More formally, let $\mathbb{N}^\mathbb{N}$ denote the space of sample paths with the standard probability induces by the stochastic matrix $P=(p(n,m))_{n,m=0}^\infty$ (considering the initial distribution as the Dirac-distribution centered at $0$). I'm looking for conditions on the coefficients $\lambda_n$ and $\mu_n$ which imply that for a.e. sample path $(x_n)_{n=0}^\infty\in\mathbb{N}^\mathbb{N}$, there exists a natural number $N$ such that $x_{N+k}=x_N+k$ for all $k\geq 0$ (that means that there are no deaths after time $N$).
Such a condition would probably be of the form "If $\mu_n$ decays [very fast], then almost every sample path has only finitely many deaths". This is actually equivalent to for almost every sample path $(x_n)_{n=0}^\infty$, the limit $\lim_{n\to\infty} (n-x_n)$ existing (note that $n-x_n$ is simply twice the number of deaths that ocurred between times $0$ and $n$).

This is what I have done so far: We can to calculate the probability of a sample path having only finitely many deaths. More precisely,


*

*The probability of a sample path having only $1$ death is $\lambda_0\lambda_1\lambda_2\cdots=\prod_{j=1}^\infty \lambda_j$ (the only path is $(0,1,2,3,\ldots)$, and $\lambda_0=1$)

*If a sample path has only one death at time $n$, then it is the path $(0,\ldots,n,n-1,n,n+1,n+2,\ldots)$, and the chance of it occuring is $\lambda_0\cdots\lambda_{n-1}\mu_n\lambda_{n-1}\lambda_n\lambda_{n+1}\cdots=\mu_n\lambda_{n-1}(\prod_{j=1}^\infty\lambda_j)$. Thus, the probability of a sample path having precisely one death is $(\prod_{j=1}^\infty\lambda_j)(\sum_{n=1}^\infty\mu_n\lambda_{n-1})$.
(The argument in the next paragraph is oversimplified, but I'm pretty sure that this is true. This becomes clearer if we calculate the probability of a sample path having 2 deaths, but the argument would be wuite long).
Note, from 2. above, that a death at time $n$ affects the probability by a term of the form $\mu_n\lambda_{n-1}$. If a sample path $(x_0=0,1=x_1,x_2,\ldots)$ has $k$ deaths at times $n_1,n_2,\ldots,n_k$, in that order, then after the time $n_i$, $x_{n_{i+1}}$ will be at least $x_{n_{i}}-1$. Hence the probability of a sample path having precisely $k$ deaths is
$$(\prod_{j=1}^\infty\lambda_j)\sum\left\{\mu_{n_1}\lambda_{n_1-1}\cdots\mu_{n_k}\lambda_{n_k-1}:n_i\geq 1,\ n_{i+1}\geq n_i-1\right\}.$$
Therefore, the probability of a path having only finitely (including $0$) many deaths is
$$(\prod_{j=1}^\infty\lambda_j)\left(1+\sum\left\{\mu_{n_1}\lambda_{n_1-1}\cdots\mu_{n_k}\lambda_{n_k-1}:k\geq 1,n_i\geq 1,\ n_{i+1}\geq n_i-1\right\}\right).$$
Thus, a.e. sample path has only finitely many deaths iff the ugly guy up here equals $1$. In particular $0<\prod_{j=1}^\infty\lambda_j=\prod_{j=1}^\infty(1-\mu_j)$, and if I remember correctly, this is equivalent to $\sum\mu_j<\infty$ (simply take $\log$'s, etc...).
I hope someone has a nicer condition for we almost always having only finitely many deaths.
 A: An elementary argument we can make is as follows: Define by $d_n$ the expected number of deaths occurring before we reach state $n+1$ given that we start in state $n$. We clearly have $d_0=0$. Further we can write an expression for $d_n$ noting that, with, probability $\lambda_n$, we will have no deaths before reaching state $n+1$, and with probability $\mu_n$ we will have an additional death, regress to state $n-1$, expecting $d_{n-1}$ deaths before returning to state $n$ for $d_n$ more expected deaths before state $n+1$:
$$d_n=\mu_n(d_{n-1}+d_n+1)$$
$$d_n=\frac{\mu}{1-\mu}(d_{n-1}+1).$$
It is quite clear that the expected number of deaths, in general, is $D=\sum_{n=0}^{\infty }d_n$, since we expect $d_0$ deaths before reaching state $1$, and $d_1$ additional deaths before reaching state $2$ and $d_2$ deaths before reaching state $3$ and so on. If $D$ is finite, the probability of a given sample having only finitely many deaths must be $1$, since if, with positive probability, infinite deaths might happen, $D$ would have to diverge.
Note that it is obvious that $\mu_n\rightarrow 0$ when $n\rightarrow\infty$. We can use this crude, but clearly necessary condition to find a sufficient condition. In particular, choose some $N$ such that for all $n>N$ it holds that $\mu_n<\frac{1}3$. Then, we have that
$$d_n<\frac{1}2(d_{n-1}+1)$$
which, in particular, implies that, as the series of $d_n$ is converging exponentially towards $1$, that for large enough $N'>N$ it will hold that, for all $n>N'$ we have
$$0< d_{n-1}<2$$
which implies moreover that
$$\frac{\mu_n}{1-\mu_n}<d_n<\frac{3\mu_n}{1-\mu_n}$$
and since we have $0<\mu_n<\frac{1}3$ in this range, we can bound the ratios:
$$\mu_n < d_n < \frac{9}2\mu_n$$
which, being linear bounds means that $D=\sum_{i=0}^{\infty}d_n$ converges if and only if $\sum_{i=0}^{\infty}\mu_n$  does. Since $D$ existing implies there being finitely many deaths in almost all cases, this means that $\sum_{i=0}^{\infty}\mu_n$ converging to a finite value is a sufficient condition. (An argument that this is also necessary is also provided below)

Also, to flesh out the comments, because knowing about the Kolmogorov zero-one law is a plus, to solve this using that, notice that the event "there are only finitely many deaths" is a tail event since (over the space of unbounded paths; the set of bounded paths obviously has measure $0$, so we ignore it) it can be defined based on the series of independent events $X_n$ defined as "there is a transition $n$ to $n-1$ at some point" and changing finitely many $X_n$ does not change whether there were only finitely many deaths. The zero-one law states that this means that the probability of there being finitely many deaths is either $1$ or $0$.
We can note that the probability of there being only finitely many deaths is at least the probability of there being $0$ deaths, which is $$\prod_{n=0}^{\infty} (1-\mu_n)$$
which is greater than $0$ if and only if $\sum_{n=0}^{\infty}\mu_n$ converges. In that case, the probability of there being finitely many deaths is positive, and hence equals $1$, by the zero-one law. Conversely, suppose the probability of there being $0$ deaths is $0$ - meaning the sum of the $\mu_n$ diverges. This implies that the probability of there being $0$ deaths, starting at a population of $n$ is also $0$ - which in turn, implies that the probability of an additional death occurring eventually, regardless of the starting sequence, is $1$, so the probability of exactly $n$ deaths occurring is $0$ for any $n$ and, "finitely many" being a countable union of "none", "exactly one", "exactly two", $\ldots$, each having probability $0$, meaning there are almost always infinitely many deaths in this case.
