Define the relation $R_n$ with $n\ge3$ by $$aR_nb\iff a\le b^n$$ This relation is not transitive because for $a\gt2$ we have $$aR(a-1)\iff a\le(a-1)^n\\\\(a-1)R(\sqrt[n]{a-1})^n)\iff(a-1)\le(\sqrt[n]{a-1})^n=(a-1)$$ But it is clear that $a$ is not in relation with $(\sqrt[n]{a-1})^n=(a-1)$ because $a\
ot\le(a-1)$.