Inequality with continued fractions: $\theta_r \geq a_{r+2}\theta_{r+1} + \theta_{r+1}$

Question

I want to prove that the following inequality is true (or that is false, I don not know but I think it is true). $$\theta_r \geq a_{r+2}\theta_{r+1} + \theta_{r+1}.$$ Here the notation is as follow: $\alpha$ is a positive irrational number. The continued fraction expansion of $\alpha$ is $$\alpha = [a_0;a_1,a_2,\ldots,a_n,\ldots].$$ In other words \begin{equation*} \alpha = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \ldots}}. \end{equation*} We denote with $\frac{p_i}{q_i}$ the irreducible fraction representing the $i$-th convergent of $\alpha$. Finally $\theta_n$ is defined as follows $$\theta_n = \vert q_n\alpha - p_n\vert.$$ Additional details: we suppose $r \geq 2$. If the inequality is false, which is the first integer $k$ for which the following weaker inequality holds? $$\theta_r \geq a_{r+2}\theta_{r+1} + \theta_{r+k+1}.$$

Answer 1

The opposite inequality is true. Since $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$, we have $$ p_n-\alpha q_n = a_n(p_{n-1}-\alpha q_{n-1}) + (p_{n-2}-\alpha q_{n-2}). $$ Since consecutive convergents alternate sides around $\alpha$, it follows that $$ \theta_n=-a_n\theta_{n-1}+\theta_{n-2}. $$ Finally, since the $\theta_n$'s are nonnegative and decreasing, we have $$ a_n\theta_{n-1}\le\theta_{n-2}<(a_n+1)\theta_{n-1}. $$ Hence the stated inequality is always false, and the modified inequality at the end of the question is actually an equality for $k=1$, and a strict inequality for $k>1$.