Evaluating $\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots} } }$

Question

What is the value of the fraction $\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots} } } $?

I let $x=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots} } } $ and hence $x=\frac{2}{3-x} $ which gives $x=1$ or $x=2$.

If I define a recurrence relation $x_n=\frac{2}{3-x_{n-1}}$ for $n\ge 1$, I will have $\lim_{n \to \infty}x_n=1$ if $x_0\neq 2,3$ and $\lim_{n \to \infty}x_n=2$ if $x_0= 2$.

Any good explanation?

Answer 1

Any good explanation?

The simplest one: draw the graph of the function one iterates, here this is the function $f:x\mapsto2/(3-x)$, on $[0,3)$. The function $f$ is increasing and its graph is above the diagonal on $[0,1)$ and on $(2,3)$, below the diagonal on $(1,2)$ and it crosses the diagonal at $1$ and at $2$. Hence the sequence $(x_n)_{n\geqslant0}$ starting from $x_0\geqslant0$ is:

• increasing to the limit point $1$ if $x_0$ is in $[0,1)$,
• constant and equal to $1$ if $x_0=1$,
• decreasing to the limit point $1$ if $x_0$ is in $(1,2)$,
• constant and equal to $2$ if $x_0=2$,
• increasing to some point $x_n\geqslant3$ if $x_0$ is in $(2,3)$, then:
  • either $x_n=3$, then $x_{n+1}$ is undefined,
  • or $x_n\gt3$, then $x_{n+1}\lt0$, then $x_{n+2}$ is in $(0,\frac23]$ hence, from time $n+2$, one is back to the first case, that is, the sequence $(x_{n+1+k})_{k\geqslant0}$ is increasing to the limit point $1$.

What is the relevant starting point to attribute a value to the continued fraction you are interested in is a debatable matter. A common choice is $x_0=0$, in which case the value of the continued fraction in the question is $\lim\limits_{n\to\infty}x_n=1$. This choice corresponds to the sequence starting with the values $$ \color{red}{\bf0},\quad \frac2{3-\color{red}{\bf0}},\quad \frac2{3-\frac2{3-\color{red}{\bf0}}},\quad \frac2{3-\frac2{3-\frac2{3-\color{red}{\bf0}}}}, $$ or, as it happens, any nonnegative starting point except $2$. The limit point $2$ is obtained when one considers the sequence starting with the values $$ \color{blue}{\bf2},\quad \frac2{3-\color{blue}{\bf2}},\quad \frac2{3-\frac2{3-\color{blue}{\bf2}}},\quad \frac2{3-\frac2{3-\frac2{3-\color{blue}{\bf2}}}}. $$ Both these arguments point to the value $1$, not $2$.

Answer 2

It depends on how you interpret the "$\cdots$" of the continued fraction. Does it mean replace the final $3$ with $3-{2\over3}$, as in

$${2\over3},\qquad{2\over3-\displaystyle{2\over3}},\qquad{2\over3-\displaystyle{2\over3-\displaystyle{2\over3}}},\qquad\text{etc.}$$

or does it mean replace the final $2$ with $2\over3-2$, as in

$$2,\qquad{2\over3-2},\qquad{2\over3-\displaystyle{2\over3-2}},\qquad{2\over3-\displaystyle{2\over3-\displaystyle{2\over3-2}}},\qquad\text{etc.}$$

You can convince yourself that the first of these is always less than $1$, and might be able to see that it's increasing and converges to $1$. The second is easily seen to always equal $2$.