1
$\begingroup$

I have encountered this question while working on my research in model theory. So do not hesitate to give full answer as it is not homework and it is just some thing small I have to check..

Let $\alpha$ be an irrational number in $(3/4,1)$. Define $k_{\alpha}(n)$ as the best rational approximation of $1/\alpha$ from below with denominator lesser or equal to $n$. i.e $$k_{\alpha}(n)=\sup\{\frac{a_n}{b_n}\in\mathbb{Q}\mid\frac{a_n}{b_n}\le 1/\alpha,b_{n}\le n \}$$ Now the Index of an irrational number $\alpha$ is defined as $$i\left(\alpha\right)=\sum_{n=1}^{\infty}\left(1-\alpha k_{\alpha}\left(n\right)\right)$$ Now assume that $\alpha$ has an infinite index, i.e $i(\alpha)=\infty$ and denote $k_{\alpha}(n)=\frac{a_n}{b_n}$ then we have that $$\lim_{n\to\infty}(\frac{1}{\alpha}-\frac{a_n}{b_n})=0$$ Does the sequence $({b_n}-\alpha\cdot a_n)$ converge? I am trying to show that it does.

I know that for any two sequences it is not enough for the ratio to converge but perhaps under these assumptions it might converge. I thought to look at the expression $(\frac{1}{\alpha}-\frac{a_n}{b_n})\cdot b_n$ and try to bound the growth of $b_n$ as we know it does not increase very fast. But it is not relay my area and Im kind of stuck... Any help will be very appreciated. Also is the Dirichlet's approximation theorem is applicable to the sequence $k_{\alpha}(n)$ or does it has the properties of Diophantine approximations?

Also if any one has suggestions for additional tags I will be happy to add them!

Thank you!

$\endgroup$
1
$\begingroup$

The natural tool to answer this question is (generalised, since $\frac{1}{\alpha} > 1$) Farey sequences.

For every $n$, there are two fractions in $\mathcal{F}_n$ that are closest to $\frac{1}{\alpha}$, let them be $$\frac{a_n}{b_n} < \frac{1}{\alpha} < \frac{c_n}{d_n}\,. \tag{1}$$ Here $k_{\alpha}(n) = \frac{a_n}{b_n}$, conforming to your notation.

By general properties of Farey sequences we have $$\frac{c_n}{d_n} - \frac{a_n}{b_n} = \frac{1}{b_nd_n} \tag{2}$$ and therefore $$0 < \frac{1}{\alpha} - \frac{a_n}{b_n} < \frac{1}{b_nd_n}\,,$$ which yields $$0 < b_n - \alpha\cdot a_n < \frac{\alpha}{d_n} \tag{3}$$ upon multiplication by $\alpha b_n$.

Thus the — almost obvious — fact that $d_n \to \infty$ implies $$\lim_{n \to \infty} (b_n - \alpha \cdot a_n) = 0\,,$$ regardless of whether $i(\alpha)$ is finite or infinite.

$\endgroup$
3
  • $\begingroup$ So $k_{\alpha}(n)$ and $\frac{c_{n}}{d_{n}}$ are neighboring terms in the Farey sequence of order $n$ and therefor we have (2)? $\endgroup$ – sha May 21 '20 at 16:10
  • $\begingroup$ Correct. And that bound yields the desired conclusion. $\endgroup$ – Daniel Fischer May 21 '20 at 16:12
  • $\begingroup$ Thanks! I am now a step closer to proving what I want. $\endgroup$ – sha May 21 '20 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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