Establishing an inequality using principal convergents and continued fraction representation.

Question

If $\theta$ is irrational with continued fraction representation $[0;a_1,a_2,\ldots]$, $\lbrace \frac{m_k}{n_k} \rbrace$ is the sequence of principal convergents of $\theta$ and $\lbrace b_k\rbrace$ is a sequence of even integers such that $\vert b_k \vert \le \vert a_{k+1} \vert$, how do we get

$$\vert b_k \vert \Vert n_k \theta \Vert \le\frac{a_{k+1}}{n_{k+1}}?$$

I have the recurrence relation $n_{k+1}=a_{k+1}n_k+n_{k-1}$.

Here's what I have: (1) $\frac{a_{k+1}}{n_{k+1}}=\frac{a_{k+1}}{a_{k+1}n_k+n_{k-1}}=\frac{1}{n_k}-\frac{n_{k-1}}{n_k}\cdot n_{k+1},$ but I'm not sure how to involve $\theta.$

Answer 1

I assume that $||x||$ is the distance between $x$ and the nearest integer. $\theta$ is between the two successive convergents $m_k/n_k$ and $m_{k+1}/n_{k+1}$, and $$\frac{m_k}{n_k}-\frac{m_{k+1}}{n_{k+1}}=\pm \frac{1}{n_k n_{k+1}}.\qquad (*)$$ Therefore, $$||n_k \theta||\le |m_k - n_k\theta|\le |m_k-n_k\frac{m_{k+1}}{n_{k+1}}|=\frac{1}{n_{k+1}}. $$ The desired result follows immediately.

(*) is standard and follows by induction from the recurrences $$ n_{k+1}=a_{k+1} n_k + n_{k-1}, \qquad m_{k+1} = a_{k+1} m_k + m_{k-1}. $$

Re the comment, if you write $$ \phi_k(x):=a_0+1/(a_1+1/(a_2+1/(\dots+1/(a_k+x)\dots), $$ then $\phi_k$ is monotonic, $\phi_k(0)=m_k/n_k$, $\phi_k(a_{k+1}^{-1})=m_{k+1}/n_{k+1}$, and $\phi_k(\zeta^{-1})=\theta$, where $$ \zeta = a_{k+1}+1/(a_{k+2}+1/(a_{k+3}+\dots>a_{k+1}. $$ Therefore, $0<\zeta^{-1}<a_{k+1}^{-1}$, so by the monotonicity of $\phi_k$, $\theta$ is between $m_k/n_k$ and $m_{k+1}/n_{k+1}$.