[proofplan] We prove both directions using the convergents of the continued fraction expansion of $\theta$. If the partial quotients are bounded, the standard error formula for convergents gives a uniform lower bound on the error of every convergent, and Legendre's criterion reduces the non-convergent rationals to the same bound. Conversely, if the partial quotients are unbounded, the upper error estimate for convergents produces rational approximations whose errors are smaller than any fixed multiple of $q^{-2}$, contradicting bad approximability. [/proofplan]

[step:Introduce the convergents and their error estimates]For each integer $n \geq 0$, let $p_n/q_n$ denote the $n$-th convergent of $\theta$, defined by the recurrences \begin{align*} p_{-2} = 0, \quad p_{-1} = 1, \quad p_n = a_n p_{n-1} + p_{n-2}. \end{align*} The denominator sequence is defined by \begin{align*} q_{-2} = 1, \quad q_{-1} = 0, \quad q_n = a_n q_{n-1} + q_{n-2}. \end{align*} Thus $p_n \in \mathbb{Z}$, $q_n \in \mathbb{N}$, and $\gcd(p_n,q_n)=1$ for every $n \geq 0$. For each $n \geq 0$, define the complete quotient $\alpha_{n+1} \in (1,\infty)$ by \begin{align*} \alpha_{n+1} = [a_{n+1};a_{n+2},a_{n+3},\dots]. \end{align*} The standard continued fraction identity gives \begin{align*} \theta = \frac{p_n\alpha_{n+1}+p_{n-1}}{q_n\alpha_{n+1}+q_{n-1}}. \end{align*} Since \begin{align*} p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}, \end{align*} we obtain \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \left| \frac{p_n\alpha_{n+1}+p_{n-1}}{q_n\alpha_{n+1}+q_{n-1}} - \frac{p_n}{q_n} \right| = \frac{1}{q_n(q_n\alpha_{n+1}+q_{n-1})}. \end{align*} Because \begin{align*} \alpha_{n+1} = a_{n+1}+\frac{1}{\alpha_{n+2}}, \end{align*} we have \begin{align*} q_n\alpha_{n+1}+q_{n-1} = q_{n+1}+\frac{q_n}{\alpha_{n+2}}. \end{align*} Since $\alpha_{n+2}>1$, this implies \begin{align*} q_{n+1} < q_n\alpha_{n+1}+q_{n-1} < q_{n+1}+q_n. \end{align*} Therefore, for every $n \geq 0$, \begin{align*} \frac{1}{q_n(q_n+q_{n+1})} < \left|\theta-\frac{p_n}{q_n}\right| < \frac{1}{q_n q_{n+1}}. \end{align*}[/step]

[guided]The convergents are the rational approximations naturally attached to the continued fraction of $\theta$. We define them recursively because the denominator recurrence is the place where the partial quotients enter the proof. For $n \geq 0$, the $n$-th convergent $p_n/q_n$ is determined by \begin{align*} p_{-2} = 0, \quad p_{-1} = 1, \quad p_n = a_n p_{n-1} + p_{n-2}. \end{align*} The denominator sequence is determined by \begin{align*} q_{-2} = 1, \quad q_{-1} = 0, \quad q_n = a_n q_{n-1} + q_{n-2}. \end{align*} The determinant identity \begin{align*} p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1} \end{align*} follows by induction from these recurrences. In particular, $p_n$ and $q_n$ are coprime, so $p_n/q_n$ is reduced. Now define the complete quotient \begin{align*} \alpha_{n+1} = [a_{n+1};a_{n+2},a_{n+3},\dots]. \end{align*} The continued fraction expansion splits after the $n$-th stage as \begin{align*} \theta = \frac{p_n\alpha_{n+1}+p_{n-1}}{q_n\alpha_{n+1}+q_{n-1}}. \end{align*} This identity lets us compute the approximation error exactly: \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \left| \frac{p_n\alpha_{n+1}+p_{n-1}}{q_n\alpha_{n+1}+q_{n-1}} - \frac{p_n}{q_n} \right|. \end{align*} Putting the two rational expressions over the common denominator gives \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \left| \frac{q_n(p_n\alpha_{n+1}+p_{n-1})-p_n(q_n\alpha_{n+1}+q_{n-1})}{q_n(q_n\alpha_{n+1}+q_{n-1})} \right|. \end{align*} Cancelling the $q_n p_n\alpha_{n+1}$ terms gives \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \left| \frac{q_n p_{n-1}-p_n q_{n-1}}{q_n(q_n\alpha_{n+1}+q_{n-1})} \right| = \frac{1}{q_n(q_n\alpha_{n+1}+q_{n-1})}. \end{align*} The last equality uses the determinant identity. The next goal is to compare the denominator $q_n\alpha_{n+1}+q_{n-1}$ with $q_{n+1}$. Since \begin{align*} \alpha_{n+1} = a_{n+1}+\frac{1}{\alpha_{n+2}}, \end{align*} we get \begin{align*} q_n\alpha_{n+1}+q_{n-1} = a_{n+1}q_n+q_{n-1}+\frac{q_n}{\alpha_{n+2}} = q_{n+1}+\frac{q_n}{\alpha_{n+2}}. \end{align*} Because $\alpha_{n+2}>1$, the extra term satisfies \begin{align*} 0 < \frac{q_n}{\alpha_{n+2}} < q_n. \end{align*} Hence \begin{align*} q_{n+1} < q_n\alpha_{n+1}+q_{n-1} < q_{n+1}+q_n. \end{align*} Taking reciprocals in the exact error formula gives \begin{align*} \frac{1}{q_n(q_n+q_{n+1})} < \left|\theta-\frac{p_n}{q_n}\right| < \frac{1}{q_n q_{n+1}}. \end{align*} This pair of inequalities is the quantitative bridge between partial quotients and Diophantine approximation.[/guided]

[step:Use bounded partial quotients to bound every convergent from below] Assume that the partial quotients are bounded. Choose $M \in \mathbb{N}$ such that \begin{align*} a_n \leq M \end{align*} for every $n \geq 1$. Since \begin{align*} q_{n+1}=a_{n+1}q_n+q_{n-1} \end{align*} and $0 \leq q_{n-1} \leq q_n$ for every $n \geq 0$, we have \begin{align*} q_{n+1} \leq (M+1)q_n. \end{align*} Therefore \begin{align*} q_n+q_{n+1} \leq (M+2)q_n. \end{align*} Using the lower convergent estimate from the previous step, \begin{align*} \left|\theta-\frac{p_n}{q_n}\right| > \frac{1}{q_n(q_n+q_{n+1})} \geq \frac{1}{(M+2)q_n^2} \end{align*} for every $n \geq 0$. [/step]

[step:Extend the lower bound from convergents to all reduced rationals] We use Legendre's criterion for [continued fractions](/page/Continued%20Fractions): if $p \in \mathbb{Z}$, $q \in \mathbb{N}$, $\gcd(p,q)=1$, and \begin{align*} \left|\theta-\frac{p}{q}\right| < \frac{1}{2q^2}, \end{align*} then $p/q$ is a convergent of $\theta$. Here this is used as a standard continued-fraction result not yet linked in the wiki: Legendre's criterion for continued fractions. Let $p \in \mathbb{Z}$ and $q \in \mathbb{N}$ satisfy $\gcd(p,q)=1$. If $p/q$ is a convergent, the previous step gives \begin{align*} \left|\theta-\frac{p}{q}\right| \geq \frac{1}{(M+2)q^2}. \end{align*} If $p/q$ is not a convergent, the contrapositive of Legendre's criterion gives \begin{align*} \left|\theta-\frac{p}{q}\right| \geq \frac{1}{2q^2}. \end{align*} Since $M \geq 1$, we have \begin{align*} \frac{1}{2} \geq \frac{1}{M+2}. \end{align*} Thus every reduced rational $p/q$ satisfies \begin{align*} \left|\theta-\frac{p}{q}\right| \geq \frac{1}{(M+2)q^2}. \end{align*} [/step]

[step:Pass from reduced rationals to arbitrary rational presentations] Let $p \in \mathbb{Z}$ and $q \in \mathbb{N}$ be arbitrary. Choose integers $p' \in \mathbb{Z}$ and $q' \in \mathbb{N}$ such that $\gcd(p',q')=1$ and \begin{align*} \frac{p}{q}=\frac{p'}{q'}. \end{align*} Then $q' \leq q$. Applying the reduced-rational estimate from the previous step gives \begin{align*} \left|\theta-\frac{p}{q}\right| = \left|\theta-\frac{p'}{q'}\right| \geq \frac{1}{(M+2)(q')^2} \geq \frac{1}{(M+2)q^2}. \end{align*} Therefore $\theta$ is badly approximable with admissible constant \begin{align*} c(\theta)=\frac{1}{M+2}. \end{align*} [/step]

[step:Use unbounded partial quotients to contradict any uniform lower constant] Conversely, assume that the partial quotients $(a_n)_{n \geq 1}$ are unbounded. Let $c>0$ be arbitrary. Since the sequence is unbounded, choose $n \geq 0$ such that \begin{align*} a_{n+1} > \frac{1}{c}. \end{align*} The denominator recurrence gives \begin{align*} q_{n+1}=a_{n+1}q_n+q_{n-1}\geq a_{n+1}q_n. \end{align*} Using the upper convergent estimate from the first step, we obtain \begin{align*} \left|\theta-\frac{p_n}{q_n}\right| < \frac{1}{q_n q_{n+1}} \leq \frac{1}{a_{n+1}q_n^2} < \frac{c}{q_n^2}. \end{align*} Thus no positive constant $c$ can satisfy \begin{align*} \left|\theta-\frac{p}{q}\right| \geq \frac{c}{q^2} \end{align*} for every $p \in \mathbb{Z}$ and $q \in \mathbb{N}$. Hence $\theta$ is not badly approximable. Combining this with the bounded-partial-quotient direction proves that $\theta$ is badly approximable if and only if $(a_n)_{n \geq 1}$ is bounded. [/step]