[proofplan] The argument has two ingredients: (1) a quantitative gap between consecutive convergents, computed from the determinant identity $p_n q_{n-1} - p_{n-1} q_n = (-1)^{n-1}$; and (2) the geometric fact that $\theta$ lies strictly between consecutive convergents, which follows from the interpolation formula for $[a_0, \ldots, a_n, \theta_{n+1}]$ applied with $\theta_{n+1} > 1$. Combining these gives $|\theta - p_n/q_n| < 1/(q_n q_{n+1})$. The denominators $q_n$ grow without bound (they are strictly increasing positive integers for $n \geq 1$), so the right-hand side tends to zero and $p_n/q_n \to \theta$. [/proofplan] [step:Compute the gap between consecutive convergents] We show \begin{align*} \left| \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} \right| &= \frac{1}{q_n q_{n+1}} \qquad (n \geq 0). \end{align*} Bringing the two fractions to a common denominator: \begin{align*} \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} &= \frac{p_{n+1} q_n - p_n q_{n+1}}{q_n q_{n+1}}. \end{align*} By the [Determinant Relations](/theorems/1759) applied at index $n + 1$, \begin{align*} p_{n+1} q_n - p_n q_{n+1} &= (-1)^n. \end{align*} Also $q_n \geq 1$ for all $n \geq 0$ (from the recurrence $q_n = a_n q_{n-1} + q_{n-2}$ with $a_n \geq 1$ for $n \geq 1$ and initial data $q_0 = 1$, $q_{-1} = 0$; see [Convergent Identity and Interpolation](/theorems/1758)), so $q_n q_{n+1} > 0$. Taking absolute values: \begin{align*} \left| \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} \right| &= \frac{|(-1)^n|}{q_n q_{n+1}} = \frac{1}{q_n q_{n+1}}. \end{align*} [/step] [step:Show $\theta$ lies strictly between consecutive convergents] Since $\theta \in \mathbb{R} \setminus \mathbb{Q}$, by the [Rationality Characterises Termination](/theorems/1757) theorem its continued fraction expansion does not terminate, so all complete quotients $\theta_0, \theta_1, \theta_2, \ldots$ are defined. By the construction of the algorithm, for each $n \geq 0$, \begin{align*} \theta &= [a_0, a_1, \ldots, a_n, \theta_{n+1}]. \end{align*} The complete quotient satisfies $\theta_{n+1} > 1 > 0$ (see Step 1 of the proof of [Rationality Characterises Termination](/theorems/1757)). By part (ii) of the [Convergent Identity and Interpolation](/theorems/1758) theorem, applied with $\beta = \theta_{n+1} > 0$, \begin{align*} \theta &= \frac{\theta_{n+1} p_n + p_{n-1}}{\theta_{n+1} q_n + q_{n-1}}, \end{align*} and this value lies strictly between $p_{n-1}/q_{n-1}$ and $p_n/q_n$ (for $n \geq 1$, where both denominators are positive integers). In particular, for $n \geq 0$, the same argument with $n$ replaced by $n + 1$ shows that $\theta$ lies strictly between $p_n/q_n$ and $p_{n+1}/q_{n+1}$: \begin{align*} \theta = [a_0, \ldots, a_{n+1}, \theta_{n+2}] = \frac{\theta_{n+2} p_{n+1} + p_n}{\theta_{n+2} q_{n+1} + q_n} \end{align*} lies strictly between $p_n/q_n$ and $p_{n+1}/q_{n+1}$ by the interpolation theorem applied with parameter $\beta = \theta_{n+2} > 0$ at level $n + 1$. [/step] [step:Bound $|\theta - p_n/q_n|$ by the gap from Step 1] Fix $n \geq 0$. Since $\theta$ lies strictly between $p_n/q_n$ and $p_{n+1}/q_{n+1}$ (Step 2), \begin{align*} \left| \theta - \frac{p_n}{q_n} \right| &< \left| \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} \right| = \frac{1}{q_n q_{n+1}}, \end{align*} using the gap computation from Step 1 in the equality. [guided] We have two key facts available: from Step 1, the two consecutive convergents $p_n/q_n$ and $p_{n+1}/q_{n+1}$ are at distance exactly $1/(q_n q_{n+1})$; from Step 2, $\theta$ lies strictly between them. When a point lies strictly between two points at distance $d$, its distance to each endpoint is strictly less than $d$. Formally: if $A$, $B$, $C$ are real numbers with $C$ strictly between $A$ and $B$, then $|C - A| < |B - A|$. This is immediate from the case analysis $A < C < B$ (giving $|C - A| = C - A < B - A = |B - A|$) and $B < C < A$ (giving $|C - A| = A - C < A - B = |B - A|$). Applying this with $A = p_n/q_n$, $B = p_{n+1}/q_{n+1}$, $C = \theta$: \begin{align*} \left| \theta - \frac{p_n}{q_n} \right| &< \left| \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} \right| = \frac{1}{q_n q_{n+1}}. \end{align*} [/guided] [/step] [step:Show $q_n \to \infty$ and conclude convergence] We show $q_n \geq n$ for all $n \geq 1$, which forces $q_n \to \infty$. From the recurrence $q_n = a_n q_{n-1} + q_{n-2}$ with $a_n \geq 1$ (for $n \geq 1$), we get $q_n \geq q_{n-1} + q_{n-2}$ for $n \geq 2$, and $q_1 = a_1 q_0 + q_{-1} = a_1 \geq 1$. Induction on $n$: the base case $n = 1$ gives $q_1 \geq 1$. If $q_{n-1} \geq n - 1$ and $q_{n-2} \geq \max(n - 2, 0)$ for some $n \geq 2$, then \begin{align*} q_n &\geq q_{n-1} + q_{n-2} \geq (n - 1) + 0 = n - 1 \end{align*} when $n = 2$, giving $q_2 \geq q_1 + q_0 = a_1 + 1 \geq 2$; and for $n \geq 3$, $q_n \geq (n-1) + 1 = n$ using $q_{n-2} \geq 1$. So $q_n \geq n$ for $n \geq 2$ (and $q_1 \geq 1$), giving $q_n \to \infty$. From Step 3, for all $n \geq 0$, \begin{align*} \left| \theta - \frac{p_n}{q_n} \right| &< \frac{1}{q_n q_{n+1}} \leq \frac{1}{q_n} \to 0 \quad \text{as } n \to \infty. \end{align*} Hence $p_n/q_n \to \theta$ as $n \to \infty$. This establishes both claims: the quantitative bound $|\theta - p_n/q_n| < 1/(q_n q_{n+1})$ and the convergence $p_n/q_n \to \theta$. [/step]