Suppose that there is an urn with infinite capacity. In the first day, we put one ball in the urn and remove it. In the second day, we put 2 balls and remove one randomly. in the $k^{th}$-day, we put $k$ balls and remove one randomly.
What is the probability that, eventually, we will put a ball and never remove it.
Well, first, it's not hard to see that before removing a ball, in the $k^{th}$ day there are:
$$\frac{k(k+1)}{2}-(k-1)=\frac{k(k-1)}{2}+1$$
Now, let's focus on some ball from the $N^{th}$ day, and for every $n\geq N$, let $A_n$ is the event a ball that was inserted in the $N^{th}$ day will be removed by, at most, the $n^{th}$ day. So $A_N\subseteq A_{N+1}\subseteq\cdots$ and the event that a ball from the $N^{th}$ will stay in the urn forever is:
$$B_N=\bigcap_{n=N}^\infty A_n^C$$
also, we get that:
$$P(A_{n+1}|A_n^C)=\frac{1}{\frac{n(n+1)}{2}+1}$$
Therefore, $\sum_{n=N}^\infty P(A_{n+1}|A_n^C)<\infty$
Those are just ideas, I don't really know how to proceed from here...