Let $P$ be a prime ideal of a commutative Noetherian ring $R$.
If $\text{Ass}(R/P^n)=\{P\}$ for infinitely many positive integers $n$, then is it true that $\text{Ass}(R/P^n)=\{P\}$ for all large enough $n$?
A trivial note: If $P$ is a power of a maximal ideal itself, then the claim is obvious.
Another note: In terms of symbolic power https://en.m.wikipedia.org/wiki/Symbolic_power_of_an_ideal, it is known that given an integer $n\ge 1$, $P^n=P^{(n)}$ holds if and only if $\text{Ass}(R/P^n)=\{P\}$.
I now realize that this happens to be true as a consequence of Brodmann's Theorem https://www.ams.org/journals/proc/1979-074-01/S0002-9939-1979-0521865-8/ which says that for any ideal $I$ in a commutative Noetherian ring $R$, it holds that $\text{Ass}(R/I^n)=\text{Ass}(R/I^{n+1})$ for eventually all large enough $n$. Consequently, if $\text{Ass}(R/I^n)=:S$ is the same set for infinitely many $n$, then $S=\text{Ass}(R/I^n)$ for all large enough $n\gg 0$.