Is this a property of an integral domain that is not a field? I am working on a specific problem and I've almost got it solved. To solve it, however, I need to prove one last claim (if it is even true):

Consider an integral domain $R$ that is not a field. Let $a$ be an element of $R$ that is not a unit or $0$. I want to prove that there cannot exist nonzero elements $r_n$, $n \in \mathbb{N}$, of $R$ such that
$$ r_1 a=r_2a^2=r_3a^3=\cdots= r_na^n=\cdots$$

First of all, is this even true? My intuition says it is. And am I missing something obvious?  This is the last step in a long proof, so I hope it is true!
EDIT: It turns out my proof goes through without this being true (which is good since it's false), but definitely thanks for the answers. It has improved my understanding.
 A: No. Let $R = \mathbb{Z}[\frac{1}{x}, 2x, 2x^2, 2x^3, ...]$, let $r_i = 2x^i$, and let $a = \frac{1}{x}$. 
A: Let $R = {}^\star \mathbb{Z}$ be the hyper-integers; that is, the ring of integers in a non-standard model of real analysis.
Let $H$ be an infinite hyper-integer.
Then, we can choose $a = 2$ and $r_i = 2^{H - i}$ to get a counter-example.
A: This is false. As Zhen Lin points out, $R$ can't be Noetherian, so let $R$ be my favourite non-Noetherian domain, $\{f(x)\in \Bbb{Q}[x]\mid f(0)\in \Bbb{Z}\}$, and consider $a=2$. Then $$x=2\frac{x}{2}=4\frac{x}{4}=8\frac{x}{8}=\ldots$$ but $2$ is not a unit.
A: Let $R$ be the ring of functions $[0,1]\to \mathbb R$ that are analytic in $(0,1]$ and  continuous on $[0,1]$.
Let $a$ be the function $x\mapsto x$.
It is not a unit because $a(0)=0$.
For each $n\in \mathbb N$, let 
$$r_n(x)=\begin{cases}x^{-n}e^{-\frac1x}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$
Note that $r_n$ is analytic in $(0,1]$ and is continuos also at $0$.
We have $$(a^nr_n)(x)=\begin{cases}e^{-\frac1x}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$
