Comparing efficiency between estimators Suppose that $\hat \theta_1, \hat \theta_2$ are two estimators of $\theta$. Furthermore, assume that
\begin{align}
\sqrt{n}(\hat \theta_1-\theta)\overset{d}{\to}N(0,V_1)\\
\sqrt{n}(\hat \theta_2-\theta-B)\overset{d}{\to}N(0,V_2),
\end{align}
where $V_{1}=\lim_{n\to\infty} V_{1,n}$, $V_{2}=\lim_{n\to\infty} V_{2,n}$ and $B:=B_n$ are all known (including $V_{1,n},V_{2,n}$).
Does it make sense to talk about relative efficiency between $\hat \theta_1$ and $\hat \theta_2$ by comparing $V_1$ and $V_2$ (or even $V_{1,n}$ and $V_{2,n}$) in the presence of term $B$?
For example, if I show that $V_{1,n}/n-V_{2,n}/n-B<0$, then can I say that $\hat\theta_1$ is relatively more efficient that $\hat\theta_2$?
*Based on the answer and a comment, I clarify that the term $B$ is relevant in the expression for $\hat \theta_2$. Hence it depends on $n$. $B$ itself is $o(1), n\to\infty$.

This is a reasonable question in settings where you 'know' $B$ in theory but you can't estimate it effectively, or if you have a bound rather than an actual value or whatever. For example, I've had settings a bit like this where there's a slightly misspecified model and the bias in an estimator is bounded above by the log likelihood ratio between the assumed and true models. A sensible comparison is of the mean squared errors, so $B^2+V_2$ vs $V_1$.
There are extensions of the Cramer-Rao inequality for MSE. Asymptotic efficiency results such as the local asymptotic minimax theorem carry over -- the local asymptotic minimax theorem bounds any 'bowl-shaped symmetric loss', which includes MSE.
