Infinite Continued Fractions

Theorem 2.1   Let $a_0, a_1, a_2, \ldots$ be a sequence of integers such that $a_n > 0$ for all $n\geq 1$, and for each $n\geq 0$, set $c_n = [a_0, a_1, \ldots a_n].$ Then $$\lim_{n\rightarrow \infty} c_n$$ exists.

Proof. For any $m\geq n$, the number $c_n$ is a partial convergent of $[a_0, \ldots, a_m]$. Recall from the previous lecture that the even convergents $c_{2n}$ form a strictly increasing sequence and the odd convergents $c_{2n+1}$ form a strictly decreasing sequence. Moreover, the even convergents are all $\leq c_1$ and the odd convergents are all $\geq c_0$. Hence $\alpha_0 = \lim_{n\rightarrow \infty} c_{2n}$ and $\alpha_1 = \lim_{n\rightarrow \infty} c_{2n+1}$ both exist and $\alpha_0\leq \alpha_1$. Finally, by a proposition from last time

$$\vert c_{2n} - c_{2n-1}\vert = \frac{1}{q_{2n}\cdot q_{2n-1}} \leq \frac{1}{2n(2n-1)} \rightarrow 0,$$

so $\alpha_0 = \alpha_1$. $\qedsymbol$

We define

$$[a_0, a_1, \ldots ] = \lim_{n\rightarrow \infty} c_n.$$

Example 2.2   We use PARI to illustrate the convergence of the theorem for $x=\pi$.

? a = contfrac(Pi)
%38 = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, ...]
? c = convergents(a)
%39 = [3, 22/7, 333/106, 355/113, 103993/33102, 104348/33215, ...]
? \p9                      \\ so we can see.
   realprecision = 9 significant digits
? [c[1]*1.0, c[3]*1.0, c[5]*1.0, c[7]*1.0]  \\ odd ones converge up to pi
%43 = [3.00000000, 3.14150943, 3.14159265, 3.14159265]
? [c[2]*1.0,c[4]*1.0,c[6]*1.0,c[8]*1.0]     \\ even ones swoop down on pi.
%44 = [3.14285714, 3.14159291, 3.14159265, 3.14159265]

Theorem 2.3   Let $x\in\mathbb{R}$ be a real number. Then

$$x = [a_0, a_1, a_2, \ldots],$$

where $a_0, a_1, a_2, \ldots$ is the sequence produced by the continued fraction algorithm.

Proof. If the sequence is finite then some $t_n=0$ and the result follows by Proposition 1.2. Suppose the sequence is infinite. By Proposition 1.2,

$$x = [a_0, a_1, \ldots, a_n, \frac{1}{t_n}].$$

By a proposition from the last lecture¹,

$$x = \frac{\frac{1}{t_n} p_n + p_{n-1}}{\frac{1}{t_n} q_n + q_{n-1}}.$$

Thus if $c_n = [a_0, a_1, \ldots, a_n]$, then

$$x - c_n = x - \frac{p_n}{q_n}$$

$$=\frac{\frac{1}{t_n} p_n q_n + p_{n-1} q_n - \frac{1}{t_n} p_n q_n - p_n q_{n-1}} {q_n \left(\frac{1}{t_n} q_n + q_{n-1}\right)}.$$

$$= \frac{p_{n-1} q_n - p_{n}q_{n-1}}{q_n\left(\frac{1}{t_n} q_n + q_{n-1}\right)}$$

$$= \frac{(-1)^n}{q_n\left(\frac{1}{t_n} q_n + q_{n-1}\right)}.$$

Thus

$$\vert x - c_n\vert = \frac{1}{q_n\left(\frac{1}{t_n} q_n + q_{n-1}\right)}$$

$$< \frac{1}{q_n(a_{n+1} q_n + q_{n-1})}$$

$$= \frac{1}{q_n \cdot q_{n+1}} \leq \frac{1}{n(n+1)} \rightarrow 0.$$

(In the inequality we use that $a_{n+1}$ is the integer part of $\frac{1}{t_n}$, and is hence $\leq \frac{1}{t_n}$.)

$\qedsymbol$