May 24, 2020

Continued Fractions, Part 3: Convergence

This is the third article in the continued fractions series, following Part 2.


A continued fraction \[ a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{\cdots}}} \] is a mere form constructed from an infinite sequence of numbers. We are going to describe when and how to associate a value to it.


Let $\kappa$ be a continued fraction $\{c_n\}_{n\in\N}$. We say $\kappa$ converges to a finite value if and only if there exists $\alpha\in\C$ such that \[ \lim_{n\to\infty}c_n=\alpha, \] and write $\kappa=\alpha$ for brevity. Also, we say $\kappa$ converges to $\infty$ if and only if \[ \forall r \in \R_+ \; \exists n_0 \in \N \; \forall n \in \N : n > n_0 \Longrightarrow |c_n| > r, \] where $\R_+=\{x\;|\;x\in\R\wedge x>0\}$, and write $\kappa=\infty$ for brevity. Finally, we say $\kappa$ converges in $\whC$ if and only if $\kappa$ converges to a finite value or to $\infty$.

A Basic Case

Theorem. Let $a, b$ be constants in $\C^\times$ such that $b/a^2$ is not a real number less than $-1/4$. Then, \[ a + \cfrac{b}{a + \cfrac{b}{a + \cfrac{b}{\cdots}}}=a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right) \] where the argument of the square root lies in $(-\pi/2, \pi/2]$.

Proof. The convergents $c_n=p_n/q_n$ are determined by the recurrence relation \[ \begin{aligned} p_{n+2}&=ap_{n+1}+bp_n, & p_1&=a, & p_0&=1, \\ q_{n+2}&=aq_{n+1}+bq_n, & q_1&=1, & q_0&=0. \end{aligned} \] Let $\alpha, \beta$ be the two roots of $z^2-az-b$. Then \[ \begin{aligned} p_{n+2}-\alpha p_{n+1}&=\beta(p_{n+1}-\alpha p_n)=\beta^{n+1}(p_1-\alpha p_0), \\ p_{n+2}-\beta p_{n+1} &=\alpha(p_{n+1}-\beta p_n)=\alpha^{n+1}(p_1-\beta p_0). \end{aligned} \]

When $\alpha \ne \beta$, we can eliminate $p_{n+2}$ to get \[ \begin{aligned} p_{n+1}&=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}p_1-\frac{\alpha^{n+1}\beta-\alpha\beta^{n+1}}{\alpha-\beta}p_0 \\ &=\frac{p_1-\beta p_0}{\alpha-\beta}\alpha^{n+1}-\frac{p_1-\alpha p_0}{\alpha-\beta}\beta^{n+1} \\ &=\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta}. \end{aligned} \] Similarly, we have \[ q_{n+1}=\frac{q_1-\beta q_0}{\alpha-\beta}\alpha^{n+1}-\frac{q_1-\alpha q_0}{\alpha-\beta}\beta^{n+1}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}. \] Therefore, by interpreting $0^0=1$, we obtain \[ c_n=\frac{p_n}{q_n}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha^n-\beta^n}. \] for any $n\in\N$. We can explicitly write $\alpha$ and $\beta$, without loss of generality, as \[ \alpha=a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right), \quad \beta=a\left(\frac{1}{2}-\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right). \] Noticing that $\Re\sqrt{1/4+b/a^2} > 0$, we have $|\alpha| > |\beta|$ and thus \[ \lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{1-(\beta/\alpha)^{n+1}}{1-(\beta/\alpha)^n}\alpha=\alpha. \]

When $\alpha=\beta$, we have $\alpha=a/2$ and \[ p_{n+2}-\alpha p_{n+1}=\alpha^{n+1}(p_1-\alpha p_0) \Longleftrightarrow \frac{p_{n+2}}{\alpha^{n+2}}-\frac{p_{n+1}}{\alpha^{n+1}}=\frac{p_1}{\alpha}-p_0=1, \] and thus $p_n=(n+1)\alpha^n$. Similarly, we have \[ \frac{q_{n+2}}{\alpha^{n+2}}-\frac{q_{n+1}}{\alpha^{n+1}}=\frac{q_1}{\alpha}-q_0=\frac{2}{a} \] and thus $q_n=2n\alpha^n/a$. Finally, we obtain \[ \lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{n+1}{n}\cdot\frac{a}{2}=\frac{a}{2}=a\left(\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{b}{a^2}}\right). \]


Kenichi Kondo © 2001-2020