$0<\varepsilon <1$. If $n_k$ and $a_k$ are positive integers for which $$n_{k+1}=a_{k+1}n_k+n_{k-1}$$ Let $L\in\mathbb{N}.$ If $L>a_k \ge 3$, what's the smallest upper bound I can place on $$\frac{\varepsilon n_{k-1}^2}{n_k^{2-\varepsilon}}?$$ by taking $k$ arbitrarily large? All of my upper bounds have been infinite, but I think there is hope since $n_k$ can be arbitrarily larger than $n_{k-1}$.
An attempt: I'd like to have an upper bound less than 1. $$n_{k+1}=a_{k+1}n_k+n_{k-1}$$ so $$\frac{\varepsilon n_{k-1}^2}{(a_kn_{k-1}+n_{k-2})^{2-\varepsilon}}<1 \iff \varepsilon n_{k-1}^2<(a_kn_{k-1}+n_{k-2})^{2-\varepsilon}.$$ However, $$ (a_kn_{k-1}+n_{k-2})^{2-\varepsilon}>(a_kn_{k-1})^{2-\varepsilon}+n_{k-2}^{2-\varepsilon}$$ , so it suffices to have $$(a_kn_{k-1})^{2-\varepsilon}+n_{k-2}^{2-\varepsilon}>\varepsilon n_{k-1}^2$$ The left side is at least $(3n_{k-1})^{2-\varepsilon}+n_{k-2}^{2-\varepsilon}$
Solve $n_{k+1}=a_{k+1}\cdot n_k+n_{k-1}$ for $a_{k+1},$ so that $$[1]\ \ a_{k+1}=\frac{n_{k+1}}{n_k}-\frac{n_{k-1}}{n_k}.$$ Assume for $n>N$ you have (shifting the index from your posted expression) $$\frac{n_k^2}{n_{k+1}}<1.$$ Then $n_{k+1}/n_k >n_k$, while clearly $n_{k-1}/n_k<1.$ We also know that $n_k >3n_{k-1}$ so that $n_k \to \infty$ as $k \to \infty.$
But then a look at the two terms on the right of [1] shows that $a_k \to \infty$ with $k$, against your assumption that $3 \le a_k < L.$
NOTE: The question has now been rephrased with more intricate assumptions regarding the ratio of successive $n_k$, making the above answer obsolete. Still, maybe some way exists of applying the idea of the answer to get something about a bound.