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Let $\delta >0$. Take $\theta \in [0,1]-\mathbb{Q},$ let $\lbrace \frac{m_k}{n_k}\rbrace$ be the sequence of principal convergents to $\theta$, obtained from the continued fraction representation $\theta =[0; a_1, a_2,...],$ where $\sup a_k <\infty$. How fast do the $n_k$ grow?

I have $$\log n_k =A_k (1+\delta)^k,$$

and I want to know if I can show that $A_k \searrow 0$. Treating $k, n_k$ as real variables, I have $$\lim_{k\rightarrow \infty}A_k=\lim_{k \rightarrow \infty}\frac{\log n_k}{(1+\delta )^k}.$$ I cannot go further since I don't know how fast the sequence $\lbrace n_k \rbrace$ grows. Using L'Hoptial's Rule I get $$\lim_{k \rightarrow \infty}A_k=\lim _{k \rightarrow \infty}\frac{n_k\cdot \frac{\text{d}}{\text{dx}}n_k}{(1+\delta)^k \log(1+\delta)}$$

Are there any insights as to whether $\lim A_k$ exists? Any ideas about how fast $n_k$ grows?

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They always grow at least as fast as the Fibonacci numbers, as can be seen from the recursive formula $n_0 =0$, $n_1 = 1$, $n_{k+1} = a_k n_k + n_{k-1}$. However, they can grow as fast as any prescribed sequence, by choosing a rapidly increasing sequence of coefficients $(a_k)$, so you will certainly not be able to show $A_k \to 0$, unless you have some additional assumptions.

If the sequence of coefficients $(a_k)$ is bounded by some number $M$, then you easily get by induction $n_k \le (M+1)^{k}$ (which is not best possible, but good enough), so $\log n_k \le k \log (M+1)$, and $\frac{n_k}{(1+\delta)^k} \to 0$ as $k \to \infty$.

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  • $\begingroup$ ok, I will edit. The a_k are bounded: $\sup a_k < \infty$ $\endgroup$ Nov 21, 2012 at 4:54

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