[proofplan] We introduce the complete quotients of the continued fraction and express $\theta$ exactly in terms of the $n$th convergent data and the next complete quotient. The determinant identity for consecutive convergents then converts the approximation error into an exact reciprocal formula. Finally, the recurrence for $q_{n+1}$ and the strict positivity of the fractional tail show that the denominator in this exact formula is strictly larger than $q_n q_{n+1}$. [/proofplan] [step:Define the complete quotients and record their relation to the partial quotients] For the [simple continued fraction](/page/Continued%20Fraction) expansion of $\theta$, define the complete quotient sequence $(\theta_k)_{k \geq 0}$ and the partial quotient sequence $(a_k)_{k \geq 0}$ recursively as follows. Set $\theta_0 := \theta$. Once $\theta_k$ has been defined, set $a_k := \lfloor \theta_k \rfloor$. We first verify, at the same time as the recursion, that $\theta_k-a_k$ is never zero. The base case is $\theta_0 = \theta \in \mathbb{R} \setminus \mathbb{Q}$. If $\theta_k$ is irrational, then $a_k \in \mathbb{Z}$ and $\theta_k-a_k$ is irrational; in particular, $\theta_k-a_k \neq 0$. Thus the next complete quotient is well-defined by \begin{align*} \theta_{k+1} := \frac{1}{\theta_k-a_k}. \end{align*} Since the reciprocal of a nonzero irrational real number is irrational, $\theta_{k+1}$ is irrational. By induction, every complete quotient $\theta_k$ is irrational. Since $a_k = \lfloor \theta_k \rfloor$, we also have $\theta_k - a_k \in (0,1)$ and $\theta_{k+1} > 1$. Hence, for every $k \geq 0$, \begin{align*} \theta_k = a_k + \frac{1}{\theta_{k+1}}. \end{align*} In particular, for every $n \geq 0$, \begin{align*} \theta_{n+1} = a_{n+1} + \frac{1}{\theta_{n+2}} > a_{n+1}. \end{align*} [/step] [step:Express $\theta$ through the $n$th convergent and the next complete quotient] We claim that, for every $n \geq 0$, \begin{align*} \theta = \frac{p_n \theta_{n+1} + p_{n-1}}{q_n \theta_{n+1} + q_{n-1}}. \end{align*} [guided] The point of introducing the complete quotient $\theta_{n+1}$ is that it represents the entire infinite tail after the first $n+1$ partial quotients. Thus the [continued fraction](/page/Continued%20Fraction) of $\theta$ can be written as the finite continued fraction \begin{align*} \theta = [a_0; a_1, \dots, a_n, \theta_{n+1}]. \end{align*} We verify the corresponding fractional-linear formula by induction on $n$. For $n=0$, the recurrence gives $p_0 = a_0$, $q_0 = 1$, $p_{-1} = 1$, and $q_{-1} = 0$. Therefore \begin{align*} \frac{p_0 \theta_1 + p_{-1}}{q_0 \theta_1 + q_{-1}} = \frac{a_0\theta_1 + 1}{\theta_1}. \end{align*} Also, \begin{align*} \frac{a_0\theta_1 + 1}{\theta_1} = a_0 + \frac{1}{\theta_1} = \theta_0 = \theta. \end{align*} Assume the formula holds at level $n-1$, with the tail variable $\theta_n$. Since \begin{align*} \theta_n = a_n + \frac{1}{\theta_{n+1}}, \end{align*} the induction hypothesis gives \begin{align*} \theta = \frac{p_{n-1}\theta_n + p_{n-2}}{q_{n-1}\theta_n + q_{n-2}}. \end{align*} Substituting $\theta_n = a_n + 1/\theta_{n+1}$ into this identity gives \begin{align*} \theta = \frac{p_{n-1}\left(a_n + \frac{1}{\theta_{n+1}}\right) + p_{n-2}}{q_{n-1}\left(a_n + \frac{1}{\theta_{n+1}}\right) + q_{n-2}}. \end{align*} Multiplying numerator and denominator by the positive number $\theta_{n+1}$ gives \begin{align*} \theta = \frac{(a_n p_{n-1} + p_{n-2})\theta_{n+1} + p_{n-1}}{(a_n q_{n-1} + q_{n-2})\theta_{n+1} + q_{n-1}}. \end{align*} Using the recurrences $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$, this becomes \begin{align*} \theta = \frac{p_n \theta_{n+1} + p_{n-1}}{q_n \theta_{n+1} + q_{n-1}}. \end{align*} This proves the formula for every $n \geq 0$. [/guided] [/step] [step:Use the determinant identity to obtain an exact error formula] For every $n \geq 0$, the determinant identity for consecutive convergents is \begin{align*} p_n q_{n-1} - p_{n-1} q_n = (-1)^{n+1}. \end{align*} This is equivalent to the usual alternating determinant identity and avoids a negative exponent at $n=0$. Indeed, for $n=0$ this reads \begin{align*} p_0 q_{-1} - p_{-1}q_0 = a_0 \cdot 0 - 1 \cdot 1 = -1 = (-1)^1, \end{align*} and the induction step follows from the recurrences. Substituting $p_n = a_n p_{n-1}+p_{n-2}$ and $q_n = a_n q_{n-1}+q_{n-2}$ gives \begin{align*} p_n q_{n-1} - p_{n-1}q_n = (a_n p_{n-1}+p_{n-2})q_{n-1} - p_{n-1}(a_n q_{n-1}+q_{n-2}). \end{align*} Cancelling the two terms $a_n p_{n-1}q_{n-1}$ yields \begin{align*} p_n q_{n-1} - p_{n-1}q_n = p_{n-2}q_{n-1} - p_{n-1}q_{n-2}. \end{align*} Reordering this expression gives \begin{align*} p_n q_{n-1} - p_{n-1}q_n = -(p_{n-1}q_{n-2}-p_{n-2}q_{n-1}). \end{align*} The induction hypothesis therefore gives \begin{align*} p_n q_{n-1} - p_{n-1}q_n = -(-1)^n = (-1)^{n+1}. \end{align*} Using the complete quotient formula from the previous step, we first compute \begin{align*} \theta - \frac{p_n}{q_n} = \frac{p_n \theta_{n+1} + p_{n-1}}{q_n \theta_{n+1}+q_{n-1}} - \frac{p_n}{q_n}. \end{align*} Putting the two fractions over the common denominator $q_n(q_n\theta_{n+1}+q_{n-1})$ gives \begin{align*} \theta - \frac{p_n}{q_n} = \frac{q_n(p_n \theta_{n+1}+p_{n-1}) - p_n(q_n\theta_{n+1}+q_{n-1})}{q_n(q_n\theta_{n+1}+q_{n-1})}. \end{align*} Cancelling the terms $q_n p_n\theta_{n+1}$ and $p_n q_n\theta_{n+1}$ in the numerator gives \begin{align*} \theta - \frac{p_n}{q_n} = \frac{q_n p_{n-1} - p_n q_{n-1}}{q_n(q_n\theta_{n+1}+q_{n-1})}. \end{align*} We also need the denominator to be positive before rewriting the absolute value as a positive reciprocal. The denominator sequence satisfies $q_{-1}=0$, $q_0=1$, and $q_{k+1}=a_{k+1}q_k+q_{k-1}$ with $a_{k+1}\geq 1$ for $k\geq 0$, so induction gives $q_n>0$ and $q_{n-1}\geq 0$ for every $n\geq 0$. Since $\theta_{n+1}>1$, this implies \begin{align*} q_n(q_n\theta_{n+1}+q_{n-1}) > 0. \end{align*} Taking absolute values and using \begin{align*} |q_n p_{n-1} - p_n q_{n-1}| = 1 \end{align*} gives the exact formula \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \frac{1}{q_n(q_n\theta_{n+1}+q_{n-1})}. \end{align*} [/step] [step:Compare the exact denominator with $q_nq_{n+1}$] For every $n \geq 0$, the denominator recurrence gives \begin{align*} q_{n+1} = a_{n+1}q_n + q_{n-1}. \end{align*} The preceding step already established $q_n > 0$ for every $n \geq 0$. Since $q_n > 0$ and $\theta_{n+1} > a_{n+1}$, we have \begin{align*} q_n\theta_{n+1}+q_{n-1} > q_n a_{n+1}+q_{n-1} = q_{n+1}. \end{align*} Multiplying by the positive integer $q_n$ yields \begin{align*} q_n(q_n\theta_{n+1}+q_{n-1}) > q_n q_{n+1}. \end{align*} Therefore the exact error formula gives \begin{align*} \left|\theta - \frac{p_n}{q_n}\right| = \frac{1}{q_n(q_n\theta_{n+1}+q_{n-1})} < \frac{1}{q_n q_{n+1}}. \end{align*} This proves the desired inequality for every $n \geq 0$. [/step]