Let $0<\varepsilon \ll \delta$. Fix $\delta$. For any $k_0 \in \mathbb{N}$, I can deduce that $1<\frac{\log n_k}{(1+\delta)^{k-k_0}\log n_{k_0}}<1+\frac{\log7}{\delta\log n_{k_0}}$
Therefore, I write $(*)\,\log n_{k}
I really want to have $k_0$ large enough so that $(**)\,\frac{1-\frac{\varepsilon}{\delta}}{A_\delta(1+\varepsilon)}\ge 1.$ Is this possible? My intuition says yes since $(*)$ gives us some freedom on the choice of $A_\delta$ where it can be less than $1$; I.e., since $\log n_k<(1+\delta)^{k-k_0}\log n_{k_0}$, there are positive numbers $\lambda<1$ for which $\log n_k<\lambda(1+\delta)^{k-k_0}\log n_{k_0}$.
If this is true, can I choose $k_0$ large enough so that $(*)$ and $(**)$ are both satisfied, or is that circular?