Is the probability that a sequence of natural numbers contains $0$ the limit of the probabilities that it contains $0$ in the first $k$ terms? Suppose $(\Omega,\mathcal{F},P)$ is a probability space where $\Omega$ is the set of sequences of natural numbers. With $s_n$ denoting the $n$th element of sequence $s$, let $$p:=P\{s\in\Omega: s_n=0\text{ for some }n\},$$ and for each $k\ge 1$ let $$q(k):=P\{s\in\Omega: s_n=0\text{ for some }n\le k\}.$$  Is it necessarily the case that $$\lim_{k\to\infty}q(k)=p$$?
For example (paraphrasing the problem in https://www.youtube.com/watch?v=A5-Q2GdD5xw&t=18s):
Imagine a type of organism (an “alien”) that lives for a single day and then dies in the act of producing between zero and three offspring, all four possibilities being equally likely. What is the probability $p$ that an alien eventually has no living descendants?
Equating $p$ with
$$\sum_{k=0}^3\operatorname{Probability}(k \text{ offspring}) \cdot  \operatorname{Probability}(\text{extinction given $k$ offspring})$$
shows that $p$ is either $1$ or $\sqrt{2}-1$.
Let each $q(k)$ be the probability that an alien has no living descendants $k$ days after its own lifetime. Since $q$ is increasing and bounded above by $p$ it has a limit not exceeding $p$. Using the same approach as before one gets a recurrence for $q(k+1)$ in terms of $q(k)$. Taking limits in this recurrence turns it into the equation from before, so $\lim_{k\to \infty}q(k)$ is either $1$ or $\sqrt{2}-1$. Since $\sqrt{2}-1$ is an upper bound of $q$ one has in fact that $\lim_{k\to \infty}q(k)=\sqrt{2}-1$.
So everything is QED if you just know that $\lim_{k\to\infty}q(k)=p$. Enter the original question.
(later) It’s also argued that the bogus answer p=1 is ruled out by the fact that the quotient
{population on (n+1)-th day} / {population on n-th day}
has the expected value
(0+1+2+3)/4 = 3/2 > 1
That is, the population is “expected to grow”. Is this argument valid?
 A: Yes, that is necessarily the case. More precisely:
If $(\Omega=\mathbb{N}^\mathbb{N},\mathcal{F},P)$ is a probability space such that $A:=\{s\in\Omega:\exists n\, (s_n=0)\}$ and $A_k:=\{s\in\Omega:\exists n\le k\,(s_n=0) \}$ $(k\ge 1)$ are events in $\mathcal{F},$ then $P(A)-P(A_k)\to 0$ as $k\to\infty;$ that is, in your notation, $p-q(k)\to 0.$
To see this note that $A_1\subseteq A_2\subseteq A_3\subseteq\cdots\subseteq A$ is an increasing chain with $A = \lim_{n\to\infty}A_n = {\Large\cup}_{n\ge 1}A_n.$ Now consider $B_1:=A_1$ and $B_n:=A_n\setminus A_{n-1},\, n\gt 1.$ Then ${\Large\cup}_{n\ge 1}A_n={\Large\cup}_{n\ge 1} B_n$ (a disjoint union); hence,
$P(A) = P({\Large\cup}_{n\ge 1}A_n)=P({\Large\cup}_{n\ge 1}B_n)=\sum_{n\ge 1}P(B_n)$ by $\sigma$-additivity of the probability measure.
Consequently, $P(A)-P(A_k)= P(A\setminus A_k)=P({\Large\cup}_{n\gt k}B_k)=\sum_{n\gt k}P(B_n)\to 0\text{ as ${k\to\infty}$},$ because the latter is the tail sum of a convergent series.
(This is a special case of the fact that a probability measure is continuous along
monotone sequences of events.)
