Let $X, X_{n,k}$ for $k,n\in\mathbb{N}$ denote independent random variables with values in $\mathbb{N}_0$. Define $N_0:=1$ and for $n\in\mathbb{N}$ set $$ N_n:=\begin{cases}0, & \text{ if }N_{n-1}=0\\X_{n,1}+\cdots+X_{n,N_{n-1}}, & \text{ if }N_{n-1}>0.\end{cases} $$ Find conditions on the distribution of $X$ for which the probability $$ q:=P(\exists n\in\mathbb{N}: N_n=0) $$ satisfies $q=0$.
In this thread ( Find conditions on the distribution on $X$, but what is meant by $X$?) I learned that $X$ is any random variable that has the same Distribution as $X_{1,1}=N_1$.
Now my next question is, how to find the desired condition under which it is $q=0$.
I think one maybe can start with $P(\exists n\in\mathbb{N}: N_n=0)=1-P(\forall n\in\mathbb{N}: N_n>0)$, but that's already the Point where I do not know how to continue.
Edit
Another possibility could be to start with $$ P(\exists n\in\mathbb{N}: N_n=0)=P(H(0)<\infty) $$ whereat $$ H(0)=\inf\left\{n\geq 1: N_n=0\right\}. $$ But here I do not know how to continue as well. Maybe $$ P_X(\exists n\in\mathbb{N}: N_n=0)=P_X(H(0)<\infty)=\sum_n P_X(H(0)=n)\\=\sum_n\sum_lP(H(0)=n,X=l) $$
Something is weird here... You can rewrite $q$ as follows
$$q=P(\bigcup_{n=1} \{N_n = 0\}) = \lim_{n \rightarrow \infty} P(N_n = 0)$$
Once the events $\{N_n = 0\}$ are increasing.
But, the sequence $P(N_n = 0)$ are increasing. So, if for some $n$ you have $P(N_n = 0)=\delta > 0$ then obviously $q \ge \delta$
So to start, $X$ can't be zero, I.e., $P(X_{11} = 0)=0$. But this isn't a sufficient, because $N_2$ could be zero with positive probability if the $X_{2j}$ have some chance to assume zero too. So is necessary that all $X$'s don't assume the value zero with positive probability.