Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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 $k \to \infty$.

share|cite|improve this answer
ok, I will edit. The a_k are bounded: $\sup a_k < \infty$ – The Substitute Nov 21 '12 at 4:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.