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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 $(*)\,\log n_{k}<A_\delta(1+\delta)^{k-k_0}\log n_{k_0}.$

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?

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closely related question (should have been linked in the question to avoid unnecessary duplication of efforts) –  joriki Nov 22 '12 at 1:22
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up vote 1 down vote accepted

Assuming you meant $\log n_k$ on the left of ($*$), you have $A_\delta>1$, so the expression on ($**$) is always $<1$.

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