= \ (1+\epsilon)/(x+1)$$

Then (C1) is satisfied with $J=0$ and $a=\epsilon$. Also, $\left[ X_{n+1}-X_n \right]^+ \le 1$, so (C2) would indeed hold (for any $p>0$, and with $V=1$) if we replaced $|X_{n+1}-X_n|$ by $[X_{n+1}-X_n]^+$.

On the other hand, it is straightforward to see that $\L(X_n)$ converges weakly to a stationary distribution $\pi(n)$, which is such that $\pi(n) \sim C n^{-1-\epsilon}$ as $n\to\infty$. In particular, $\sum_n n \, \pi(n) =
\infty$. It follows that $\E(X_n) \to \infty$, i.e. that $\{X_n\}$ is [*not*]{} stable in this case. We conclude that Theorem \[th:1\] does [*not*]{} continue to hold if we consider only the positive part of $X_{n+1}-X_n$ in condition (C2).

**Remark. This last counter-example only works when $a \le V^{1/p}$. In the case where $X_n \geq 0$ for all $n$, this appears to be an extremal counterexample, leading to the following open question:**

> Does Theorem \[th:1\] continues to hold for sufficiently large $a$ if we assume $X_n \geq 0$ and replace $|X_{n+1}-X_n|$ by $[X_{n+1}-X_n]^+$ in (C2) ?

Despite this counter-example, the hypotheses of Theorem \[th:1\] may indeed be weakened to allow some large negative increments. However, both condition (C1) [*and*]{} condition (C2) must be identically modified so that negative drift is still 