[proofplan] We first prove the standard determinant identity for consecutive convergents from the recurrence relations. This identity gives the exact sign and size of the difference between adjacent convergents, and also shows that even-indexed and odd-indexed convergents move monotonically in opposite directions. The two monotone subsequences are nested on opposite sides of one another, and the gap between adjacent even and odd convergents tends to zero because the denominators $q_n$ tend to infinity. Hence the even and odd subsequences have the same limit, so the full sequence converges. [/proofplan] [step:Verify positivity and divergence of the denominators] We use the standard initial values for the numerator and denominator sequences of simple continued-fraction convergents: \begin{align*} p_{-2} := 0, \qquad p_{-1} := 1, \qquad q_{-2} := 1, \qquad q_{-1} := 0. \end{align*} For every integer $n \geq 0$, define $p_n \in \mathbb{Z}$ and $q_n \in \mathbb{Z}$ recursively by \begin{align*} p_n := a_n p_{n-1} + p_{n-2}, \qquad q_n := a_n q_{n-1} + q_{n-2}. \end{align*} The $n$th convergent is the real number $p_n/q_n$ whenever $q_n \neq 0$. For $n = 0$, the recurrence gives $q_0 = a_0 q_{-1} + q_{-2} = 1$. For $n = 1$, it gives $q_1 = a_1 q_0 + q_{-1} = a_1 \geq 1$. Since $a_n \in \mathbb{Z}_{>0}$ for every $n \geq 1$, induction gives $q_n > 0$ for every $n \geq 0$. Moreover, for every $n \geq 2$, \begin{align*} q_n = a_n q_{n-1} + q_{n-2} \geq q_{n-1} + q_{n-2} > q_{n-1}. \end{align*} Thus $(q_n)_{n=1}^{\infty}$ is eventually strictly increasing and hence $q_n \to \infty$ as $n \to \infty$. [/step] [step:Prove the determinant identity for consecutive convergents] Define the integer sequence $(D_n)_{n=0}^{\infty}$ by \begin{align*} D_n := p_n q_{n-1} - p_{n-1} q_n. \end{align*} For $n = 0$, \begin{align*} D_0 = p_0 q_{-1} - p_{-1} q_0 = 0 - 1 = -1. \end{align*} For $n \geq 1$, using the recurrence relations for $p_n$ and $q_n$, \begin{align*} D_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*} Expanding and cancelling the two terms $a_n p_{n-1}q_{n-1}$ gives \begin{align*} D_n = p_{n-2}q_{n-1} - p_{n-1}q_{n-2} = -D_{n-1}. \end{align*} Therefore induction gives \begin{align*} D_n = (-1)^{n+1} \end{align*} for every $n \geq 0$. Consequently, for every $n \geq 1$, \begin{align*} \frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} = \frac{p_n q_{n-1} - p_{n-1}q_n}{q_n q_{n-1}} = \frac{(-1)^{n+1}}{q_n q_{n-1}}. \end{align*} [guided] The determinant identity measures exactly how two neighbouring convergents differ. Define \begin{align*} D_n := p_n q_{n-1} - p_{n-1}q_n. \end{align*} This is the numerator obtained after subtracting $p_{n-1}/q_{n-1}$ from $p_n/q_n$. For $n = 0$, the initial values give \begin{align*} D_0 = p_0 q_{-1} - p_{-1}q_0 = 0 - 1 = -1. \end{align*} Now let $n \geq 1$. Substituting the recurrence formulas $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$ gives \begin{align*} D_n = p_n q_{n-1} - p_{n-1}q_n. \end{align*} After substituting the recurrence relations, this becomes \begin{align*} D_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*} Expanding the products gives \begin{align*} D_n = a_n p_{n-1}q_{n-1} + p_{n-2}q_{n-1} - a_n p_{n-1}q_{n-1} - p_{n-1}q_{n-2}. \end{align*} The two terms containing $a_n p_{n-1}q_{n-1}$ cancel, so \begin{align*} D_n = p_{n-2}q_{n-1} - p_{n-1}q_{n-2} = -D_{n-1}. \end{align*} Thus the determinant alternates sign at each step, starting from $D_0 = -1$. Hence \begin{align*} D_n = (-1)^{n+1} \end{align*} for every $n \geq 0$. Since $q_n > 0$ and $q_{n-1} > 0$ for $n \geq 1$, we may divide by $q_n q_{n-1}$. This yields \begin{align*} \frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} = \frac{p_n q_{n-1} - p_{n-1}q_n}{q_n q_{n-1}} = \frac{(-1)^{n+1}}{q_n q_{n-1}}. \end{align*} This formula gives both the sign and the exact size of the jump between consecutive convergents. [/guided] [/step] [step:Show that even convergents increase and odd convergents decrease] For every $n \geq 0$, compare the convergents with indices $n$ and $n+2$. Using the recurrence relations, \begin{align*} p_{n+2}q_n - p_n q_{n+2} = (a_{n+2}p_{n+1} + p_n)q_n - p_n(a_{n+2}q_{n+1} + q_n). \end{align*} Cancelling the two terms $p_n q_n$ and factoring out $a_{n+2}$ gives \begin{align*} p_{n+2}q_n - p_n q_{n+2} = a_{n+2}(p_{n+1}q_n - p_n q_{n+1}). \end{align*} By the definition of $D_{n+1}$ and the determinant identity, \begin{align*} p_{n+2}q_n - p_n q_{n+2} = a_{n+2}D_{n+1} = a_{n+2}(-1)^n. \end{align*} Since $a_{n+2} > 0$ and $q_{n+2}q_n > 0$, it follows that \begin{align*} \frac{p_{n+2}}{q_{n+2}} - \frac{p_n}{q_n} = \frac{a_{n+2}(-1)^n}{q_{n+2}q_n}. \end{align*} Thus the even subsequence $(p_{2k}/q_{2k})_{k=0}^{\infty}$ is increasing, and the odd subsequence $(p_{2k+1}/q_{2k+1})_{k=0}^{\infty}$ is decreasing. [/step] [step:Place every even convergent below every odd convergent] For every $n \geq 1$, the adjacent difference formula gives \begin{align*} \frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} = \frac{(-1)^{n-1}}{q_n q_{n-1}}. \end{align*} In particular, if $n = 2k+1$ is odd, then \begin{align*} \frac{p_{2k+1}}{q_{2k+1}} - \frac{p_{2k}}{q_{2k}} = \frac{1}{q_{2k+1}q_{2k}} > 0. \end{align*} Hence each even convergent lies below the next odd convergent. Since the even subsequence is increasing and the odd subsequence is decreasing, for all integers $j,k \geq 0$, \begin{align*} \frac{p_{2j}}{q_{2j}} \leq \frac{p_{2\max\{j,k\}}}{q_{2\max\{j,k\}}} < \frac{p_{2\max\{j,k\}+1}}{q_{2\max\{j,k\}+1}} \leq \frac{p_{2k+1}}{q_{2k+1}}. \end{align*} Thus every even convergent is less than every odd convergent. [/step] [step:Identify the common limit of the two monotone subsequences] Define two real sequences $(e_k)_{k=0}^{\infty}$ and $(o_k)_{k=0}^{\infty}$ by \begin{align*} e_k := \frac{p_{2k}}{q_{2k}}, \qquad o_k := \frac{p_{2k+1}}{q_{2k+1}}. \end{align*} The sequence $(e_k)_{k=0}^{\infty}$ is increasing and bounded above by $o_0$, while $(o_k)_{k=0}^{\infty}$ is decreasing and bounded below by $e_0$. By the [Monotone Convergence Theorem for Real Sequences](/page/Monotone%20Convergence%20Theorem%20for%20Real%20Sequences), both subsequences converge in $\mathbb{R}$. Let \begin{align*} L_e := \lim_{k \to \infty} e_k, \qquad L_o := \lim_{k \to \infty} o_k. \end{align*} Since every even convergent is less than every odd convergent, we have $e_j \leq o_k$ for all integers $j,k \geq 0$. Taking the limit as $j \to \infty$ with $k$ fixed gives $L_e \leq o_k$ for every $k \geq 0$, and then taking the limit as $k \to \infty$ gives $L_e \leq L_o$. Since $(e_k)_{k=0}^{\infty}$ is increasing and converges to $L_e$, we have $e_k \leq L_e$ for every $k \geq 0$. Since $(o_k)_{k=0}^{\infty}$ is decreasing and converges to $L_o$, we have $L_o \leq o_k$ for every $k \geq 0$. Therefore, for every $k \geq 0$, \begin{align*} 0 \leq L_o - L_e \leq o_k - e_k. \end{align*} Using the adjacent difference formula with $n = 2k+1$, \begin{align*} o_k - e_k = \frac{p_{2k+1}}{q_{2k+1}} - \frac{p_{2k}}{q_{2k}} = \frac{1}{q_{2k+1}q_{2k}}. \end{align*} Because $q_n \to \infty$, the right-hand side tends to $0$. By the [Squeeze Theorem](/page/Squeeze%20Theorem), $L_o - L_e = 0$, so $L_e = L_o$. [guided] The two parity subsequences have already been separated: every even convergent lies below every odd convergent. Define the maps $e: \mathbb{N} \cup \{0\} \to \mathbb{R}$ and $o: \mathbb{N} \cup \{0\} \to \mathbb{R}$ by \begin{align*} e_k := \frac{p_{2k}}{q_{2k}}, \end{align*} and \begin{align*} o_k := \frac{p_{2k+1}}{q_{2k+1}}. \end{align*} The sequence $(e_k)_{k=0}^{\infty}$ is increasing and is bounded above by $o_0$, because every even convergent is below every odd convergent. By the [Monotone Convergence Theorem for Real Sequences](/page/Monotone%20Convergence%20Theorem%20for%20Real%20Sequences), $(e_k)_{k=0}^{\infty}$ converges in $\mathbb{R}$; denote its limit by $L_e \in \mathbb{R}$. Similarly, $(o_k)_{k=0}^{\infty}$ is decreasing and bounded below by $e_0$, so it converges in $\mathbb{R}$; denote its limit by $L_o \in \mathbb{R}$. Why must these two limits be the same? The nesting gives $e_j \leq o_k$ for all integers $j,k \geq 0$. Fixing $k$ and passing to the limit as $j \to \infty$ gives $L_e \leq o_k$ for every $k \geq 0$. Passing to the limit as $k \to \infty$ then gives $L_e \leq L_o$. Since an increasing convergent sequence stays below its limit and a decreasing convergent sequence stays above its limit, we also have $e_k \leq L_e$ and $L_o \leq o_k$ for every $k \geq 0$. Therefore \begin{align*} 0 \leq L_o - L_e \leq o_k - e_k. \end{align*} The adjacent difference formula at the odd index $n = 2k+1$ gives \begin{align*} o_k - e_k = \frac{p_{2k+1}}{q_{2k+1}} - \frac{p_{2k}}{q_{2k}} = \frac{1}{q_{2k+1}q_{2k}}. \end{align*} Since $q_n \to \infty$, the product $q_{2k+1}q_{2k}$ tends to $\infty$, and hence \begin{align*} \frac{1}{q_{2k+1}q_{2k}} \to 0. \end{align*} The [Squeeze Theorem](/page/Squeeze%20Theorem) for real sequences now gives $L_o - L_e = 0$. Thus $L_e = L_o$. [/guided] [/step] [step:Conclude convergence of the full sequence of convergents] Let $L := L_e = L_o$. If $n$ is even, then $n = 2k$ for some $k \geq 0$, and \begin{align*} \frac{p_n}{q_n} = e_k \to L. \end{align*} If $n$ is odd, then $n = 2k+1$ for some $k \geq 0$, and \begin{align*} \frac{p_n}{q_n} = o_k \to L. \end{align*} Thus both parity subsequences converge to the same real number $L$. Therefore the full sequence of convergents $(p_n/q_n)_{n=0}^{\infty}$ converges in $\mathbb{R}$. [/step]