Divisible totally ordered additive abelian groups

Question

Let $(G,+,\leq)$ be a divisible totally ordered additive abelian group and $g_{1},g_{2}\in G$. If for every integer $n>1$, $g_{1}\geq (1-\frac{1}{n})g_{2}$, Can we have $g_{1}\geq g_{2}$? Thanks for any answer.

Answer 1

Hint: Consider $G=\Bbb Q\times\Bbb Q$ with the lexicographic order $(a,b)\le (c,d)\iff a<c$ or $(a=c)\land (b\le d)$.
Here viewing the first coordinate as the ordinary rational numbers, $\varepsilon:=(0,1)$ will behave like an infinitesimal: it's positive but less than $\frac1n(1,0)$ for every $n\ge 1$.