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$(n_k)$ is a sequence of denominators for the sequence of prinicpal convergents of some irrational number, so $n_k \rightarrow \infty,\delta>0$. Let $0<\varepsilon \ll \delta$. I'm also given that $n_k \le (M+1)^k$.

I have deduced that for any specified natural number $k_0$ $$(*)1<\,\frac{\log n_k}{(1+\delta)^{k-k_0}\log n_{k_0}}<1+\frac{\log7}{\delta \log n_{k_0}}.$$

Therefore, by taking $k_0$ large I can have the expression in the middle tend to $1$. I'm assuming it is ok to say $$ (**)\,\log n_k=A(1+\delta)^{k-k_0}\log n_{k_0},$$

as long as I acknowledge that $A$ depends on $k, k_0,$ and $\delta$, where the latter two are predetermined.

$A\rightarrow 0$ since $$\lim_{k\rightarrow \infty}A=\lim_{k \rightarrow \infty}\frac{\log n_k}{(1+\delta)^{k-k_0}\log n_{k_0}}\le\lim_{k \rightarrow \infty}\frac{k\log(M+1)}{(1+\delta)^{k-k_0}\log n_{k_0}}=0$$

$A\rightarrow 1$ since $(*)$ implies that the ratio tends to $1$ as $k_0$ increases.

Where did I go wrong? I only want $A\rightarrow 0.$

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Why do you write "as $k_0$ increases" when the limit in the displayed equation is for $k\to\infty$? –  joriki Nov 21 '12 at 23:08
    
$A$ depends on both $k$ and $k_0$. Once $k_0$ is chosen, I justtake the limit with respect to $k$. On the other hand, $A$ tends to $1$ when I choose larger values of $k_0$. –  The Substitute Nov 21 '12 at 23:32

2 Answers 2

up vote 1 down vote accepted

In your second estimate, you deduce that $$ \frac{\log n_k}{(1+\delta)^{k-k_0}\,\log n_{k_0}}\to0 $$ as $k\to\infty$. But in (*) you say that those numbers are always above $1$. One of the two estimates is wrong. Which one it is, depends on how the numbers $n_k$ behave.

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$(*)$ doesn't imply that the ratio tends to $1$. The upper bound tends to $1$, and this is compatible with the bounded quantity tending to $0$.

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Thank you. Sorry, I've edited (*). It is also bounded below by $1$. –  The Substitute Nov 21 '12 at 23:33

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