Zero-divisors and R-sequences Here is an exercise which I can not solve. You may find it in Commutative Rings by Irving Kaplansky, p. 103, ex. 13.

Let $R$ be a commutative ring with $1$ (not necessarily Noetherian) and $A$ an $R$-module. Let $x_1,...,x_m$ be an $R$-sequence (regular sequence) on $A$, and write $I=(x_1,\dots,x_m).$
a) Prove that $Z_R(A/I^nA)=Z_R(A/IA)$ for any $n \in \mathbb{N}$, where for an $R$-module $M$, we denote $Z_R(M):= \{ r \in R \mid \exists\ 0 \not= m \in M \; \text{s.t.} \; rm=0 \}.$
b) Prove that $I^nA/I^{n+1}A$ is isomorphic to a direct sum of copies of $A/IA.$

 A: About a): prove by induction on the number of generators of $I$. We denote $A$ by $M$.
First step, $I=(x_1)$ we want to prove $Z_R(M/IM)=Z_R(M/I^nM)$.
Let $a\in Z_R(M/x_1M)$ then there exists $m\notin x_1M$  s.t. $am\in x_1M$ and $x_1^{n-1}m\notin x^nM$ (since $x_1$ is regular) thus $ax_1^{n-1}m\in x^nM$.
Conversely, $a\in Z_R(M/I^nM)$, then there exists $m\notin x_1^nM$ s.t. $am\in x_1^nM$ let $m\in x^lM$ but $m\notin x^{l+1}\cup\cdots\cup x_1^nM$. then $m=x^lm'$ ,$m'\notin x_1M$ but $am'\in x_1M$. Thus $n=1$, $Z_R(M/IM)=Z_R(M/I^nM)$.
Second, let $N=(M/(x_1,...,x_{m-1})M)$ then $Z_R(M/(x_1,...,x_{m-1},x_m)M)=Z_R(N/x_mN)=Z_R(N/x_m^nN)=Z_R(M/I^nM)$.
EDIT: Something wrong in last "=". It should be $Z_R(M/(I^n+(x_1,...,x_{m-1}))M$.

Time past, the problem is still not solved. So the update edit comes.
Update:
Lemma. $f_1,\ldots,f_r$ is an $M$-regular sequence, let $J=(f_1,\ldots,f_r)$. Then 
$$M/JM[X_1,\ldots,X_r]\cong \oplus_{n\geq 0}J^nM/J^{n+1}M.
$$
Then morphism is given by $\overline{m}X_1^{e_1}\cdots X_r^{e_r}\mapsto mf_1^{e_1}\cdots f_r^{e_r}\mod J^{e_1+\cdots+e_r+1}M$.
Proof. See lemma 66.2 on page 139 in "Commutative Algebra'' on the book of "stack project".
Now, applying this lemma we can see that the second statement is true. 
For the first assertion, we have an exact sequence
$$
0\to J^nM/J^{n+1}M\to M/J^{n+1}M\to M/J^nM\to 0.
$$
Since $J^nM/J^{n+1}M$ is a direct sum of $M/JM$, the set of nonzero divisors on $J^nM/J^{n+1}M$ is same as $M/JM$'s and also same as $M/J^nM$'s by induction. 
Hence we can see the set of nonzero divisors on $M/J^{n+1}M$ is same as $M/JM$'s by Snake lemma.
