The puzzling thing I am facing is Suppose we have two random variables $X$ and $R$ such that $E(X^{-1}R)=1$. Now let $\tilde{X}=X+\mathcal{E}$ where $\mathcal{E}=X\epsilon$ and $\epsilon \sim \mathcal{N}(0,\sigma^2)$
By Taylor expansion $\tilde{X}^{-1}=(X+\mathcal{E})^{-1}=X^{-1}-X^{-2}\mathcal{E}+2X^{-3}\mathcal{E}^2/2!-6X^{-4}\mathcal{E}^3/3!+24X^{-5}\mathcal{E}^4/4!$ ... $=X^{-1}-X^{-1}\epsilon+X^{-1}\epsilon^2-X^{-1}\epsilon^3+X^{-1}\epsilon^4$ ... using $\mathcal{E}=X\epsilon$.
Now multiplying both side of equality by random variable $R$ and taking expectations we get: $E(\tilde{X}^{-1}R)=1+E(\epsilon^2)+E(\epsilon^4)+E(\epsilon^6)+E(\epsilon^8)$...
Since every $E(X^{-1}R\epsilon^n)$ can be written as $E(X^{-1}R)E(\epsilon^n)$ by independence assumption and $E(\epsilon^n)=0$ for all odd $n$ and $E(\epsilon^n)=\sigma^n(n-1)(n-3)\cdots 3\cdot 1$ for all even $n$ it can be written:
$E((X+\mathcal{E})^{-1}R)=1+\sigma^2+3\sigma^4+15\sigma^6+105\sigma^8 ...$
And this will eventually diverge no matter how small $\sigma$ one starts with.
So we started with $E(X^{-1}R)=1$ and by introducing a very small perturbation in $X$ arrive at $E(\tilde{X}^{-1}R)=+\infty$ or am I doing something wrong.