Does the continued fraction $\frac{2}{3 - \frac{2}{3 - \ddots}}$ have two values?

Question

I was trying to prove that $$1 = \frac{2}{3 - \frac{2}{3 - \dots}}$$ It's clear that $$\begin{aligned} & & x = \frac{2}{3 - \frac{2}{3 - \dots}} \\ & \Rightarrow & x = \frac{2}{3 - x} \\ & \Rightarrow & x \in \{1, 2\} \end{aligned}$$ But I can't seem to eliminate the option $x = 2$. Also, while the seqeunce $$1 = 1, \frac{2}{3 - 1} = 1, \frac{2}{3 - \frac{2}{3 - 1}} = 1, \dots$$ clearly converges to $1$, the sequence $$2 = 2, \frac{2}{3 - 2} = 2, \frac{2}{3 - \frac{2}{3 - 2}} = 2, \dots$$ converges to $2$. I've tried using different values of $c$ in the sequence $$c,\frac{2}{3 - c}, \frac{2}{3 - \frac{2}{3 - c}}, \dots$$ and it seems that for all $c \in \mathbb{R} \setminus \{2\}$, the sequence converges to $1$ but when $c = 2$, it converges to $2$. This means that the fraction doesn't converge to a unique number and should be left undefined, right? Then why is the fraction supposedly equal to $1$? Also, are there other examples of generalized continued fractions that can converge to multiple values (maybe even an arbitrary number)?

Answer 1

The Continued Fraction

A continued fraction defines a sequence. If that continued fraction can meaningfully be given a value, that value is the limit of the sequence and must necessarily be unique.

In the case of the continued fraction described in the question above, the sequence begins with $$x_0 = \frac{2}{3},\qquad x_1 = \frac{2}{3-\frac{2}{3}},\qquad x_2 = \frac{2}{3-\frac{2}{3-\frac{2}{3}}},\qquad \dotsc.$$ Note that there are no variables running around here. The first step in assigning a value to this sequence is to show that this sequence actually converges, i.e. that $\lim_{n\to\infty} x_n$ exists. To show that this converges, it is helpful to see that $$ x_n = \frac{2}{3 - x_{n-1}}.$$ Working out the first several terms by hand (or with a tool like GeoGebra), it appears that the sequence is both bounded above by $1$, and monotonically increasing. So it is reasonable to attempt to prove these things, as this would show that the limit exists.

First, observe that $x_0 < 1$. For induction, suppose that $x_{k-1} < 1$ for some $k$. Then \begin{align} x_{k-1} < 1 &\iff 3-x_{k-1} > 3-1 = 2 \\ &\iff \frac{2}{3-x_{k-1}} < \frac{2}{2} = 1. \end{align} Therefore, by induction $x_{n} < 1$ for all $n$. This gives an upper bound of $1$ (which, by the way, rules out the possibility that the limit is $2$, as conjectured in the question).

Now observe that $$ x_n - x_{n-1} = \frac{2}{3-x_{n-1}} - x_{n-1} = \frac{2 - 3x_{n-1} + x_{n-1}^2}{3-x_{n-1}} = \frac{(x_{n-1} - 2)(x_{n-1}-1)}{3-x_{n-1}}.$$ Both the numerator and denominator of this fraction are positive when $x_{n-1} < 1$ (which has already been shown), hence $x_{n} - x_{n-1} > 0$ for all $n$. Therefore the sequence is increasing.

Every sequence which is monotonically increasing and bounded above has a limit, therefore the sequence of partial fractions being considered here has some limit, say $ \lim_{n\to\infty} x_{n} = L$. Since $x_n < 1$ for all $n$, it follows that $L \le 1$. Then, since $$ x_n = \frac{2}{3-x_{n-1}}, $$ it is possible to take limits on both sides of the equation to get $$ L = \lim_{n\to \infty} x_n = \lim_{n\to \infty} \frac{2}{3-x_n} = \frac{2}{3-L}, $$ noting that the last equality follows from continuity of the rational function. Solving for $L$ gives $$ L = \frac{2}{3-L} \iff 3L - L^2 = 2 \iff L^2 - 3L + 2 = (L-2)(L-1) = 0 \iff L \in \{1,2\}. $$ But it has already been shown that $L \le 1$, thus $L=1$.

Therefore $$\frac{2}{3-\frac{2}{3-\frac{2}{3-\dotsb}}} = 1.$$

The Dynamical System

The recurrence relation $$x_{n} = \frac{2}{3-x_{n-1}}$$ can be thought of as an interesting object by itself, taken apart from the continued fraction which originally defined it. The continued fraction in the question is the result of asking what happens as $n$ tends to $\infty$ when the initial value is $x_0 = \frac{2}{3}$. But what happens if a different initial value is chosen?

With a little more work (which I am not going to do here), it can be shown that different initial conditions give rise to different behaviours. In the notation of the question, different choices of the initial value $x_0$ correspond to the different values of $c$ in the question.

Roughly speaking, the system has only a limit number of possible behaviours:

• if $x_0 < 1$, then the sequence will be monotonically increasing, and will have limit $1$;
• if $x_0 = 1$, then the sequence will be the constant sequence in which every term is $1$;
• if $1 < x_0 < 2$, then the sequence will be monotonically decreasing, and will have limit $1$;
• if $x_0 = 2$, then the sequence will be the constant sequence in which ever term is $2$;
• if $x_0 = 3$, the sequence will be undefined (due to a division by zero right off the bat); and
• if $x_0 > 3$, then $x_1$ will be negative, after which point the sequence will be monotonically increasing with limit $1$.

For values of $x_0$ between $2$ and $3$, things are a little more compliated: for most of these values, the sequence will increase for a number of terms, eventually exceed $3$, then jump to something negative and increase monotonically to $1$. However, there are a countable number of initial values in $(2,3)$ which will eventually lead to $x_n = 3$ for some $n$, after which the sequence will be undefined. For example, if $x_0 = \frac{7}{3}$, then $$ x_1 = \frac{2}{3-\frac{7}{3}} = \frac{2}{2/3} = 3.$$ In this case, $x_2$ will be undefined.

Roughly speaking (again), this dynamical system has two equilibrium points. The value $1$ is a stable equilibrium: values "near" this point are "sucked into" it as the dynamics evolve, and if the system starts at precisely this point, it will remain at that point forever. The value $2$, on the other hand, is an unstable equilibrium: if the system starts at this value, it will remain there forever, but initial values which are "near" $2$, but not precisely $2$, will be repelled away from this value (and will, generally speaking, eventually be attracted to $1$).

There are more advanced tools which can make sense of the behavior of this dynamical system, but these are far beyond what I can fit into an answer here. In this particular case, the jargony words to Google are, perhaps, "nonlinear discrete time dynamical systems".

Answer 2

Define $a_0=\frac{2}{3}$ and $a_n=\frac{2}{3-a_{n-1}}$. The usual definition of $\frac{2}{3-\frac{2}{3-\dots}}$ is $\lim_{n\to\infty}a_n$.

One can verify by induction that $a_n$ is a strictly increasing sequence, bounded by $1$. so there exists a limit and by your calculation $\lim_{n\to\infty}\left(\frac{2}{3-a_{n-1}}=a_n\right)\Longrightarrow \frac{2}{3-a_\infty}=a_\infty \Longrightarrow a_\infty\in\{1,2\}$, where $a_\infty:=\lim_{n\to\infty}a_n$. So the limit can only be $1$, if it exists. It exists because $a_n$ is bounded and monotone, so we get $\frac{2}{3-\frac{2}{3-\dots}}=1$.

You could also define $\frac{2}{3-\frac{2}{3-\dots}}$ as $\lim_{n\to\infty}b_n$, where $b_0=\frac{2}{3-c}$ and $b_n=\frac{2}{3-b_{n-1}}$, which is what you did with your $c$ values that you put in. These however are just different limits, with possibly different values.

As you see, the expression $\frac{2}{3-\frac{2}{3-\dots}}$ has no meaning without rigourous definition and once you fix your definition, the value is determined by one limit and therefore unique if it exists. As already said the usual definition is the one with $a_n$.

Edit: With the $b_n$, you basically change the inital value, which can (and will) lead to different limits, since a recurrence relation can have multiple attractors.