how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
 A: This continued fraction is the limit of 
the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.
A: Let us define two series. The first is
\begin{align}
a_1 &= 3 \\
a_2 &= 3 - \frac{2}{3} \\
a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\
a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\
&\vdots \\
a_{n+1} &= 3 - \frac{2}{a_n} \quad (*)
\end{align}
and 
\begin{align}
b_1 &= 3 - 2 \\
b_2 &= 3 - \frac{2}{3-2} \\
b_3 &= 3 - \frac{2}{3 - \frac{2}{3-2}} \\
b_4 &= 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{3-2}}} \\
&\vdots \\
b_{n+1} &= 3 - \frac{2}{b_n} \quad (**) \\
\end{align}
Note: This is the same recurrence relation $(*)$ or $(**)$ but with different start value $a_1 = 3$ and $b_1 = 1$.
For convergence we need $a_{n+1} - a_{n} \to 0$ or $a_n \to a$.
This way (in case of convergence) equations $(*)$ and $(**)$ have a limit
$$
a = 3 - \frac{2}{a} \quad (\#)
$$
which has indeed the solutions $a = 1$ and $a = 2$.
However that means we could also try
$$
c_n = 3 - \frac{2}{c_{n+1}}
$$
or
$$
c_{n+1} = \frac{2}{3 - c_n} \quad (\#\#)
$$
because it has the limit form $(\#)$.
Note that equation $(\#\#)$ is quite different from equation $(*)$ (see image below).
And indeed this recurrence relation $(\#\#)$ works too. Using $c_1 = 1$ will give $c_n \to 1$, Using $c_1 = 2$ will give $c_n \to 2$. Using $c_1 = 1000$ will give $c_n \to 1$.
So why is this? Still two solutions and the start value decides the limit.
Here is an image:

The green graph is related to $(*)$:
$$
f(x) = 3-\frac{2}{x}
$$
the blue graph is related to $(\#\#)$:
$$
g(x) = \frac{2}{3-x}
$$ 
and the red graph is the identity function: 
$$
\mbox{id}(x) = x
$$
We see that both $f$ and $g$ hit the identity at $x=1$ and $x=2$. Those points are fixed points of $f$ and $g$:
\begin{align}
x^* &= f(x^*) \\
x^* &= g(x^*)
\end{align}
And one could now try to apply the theory of fixed points, esp. properties of fixed point iterations.
\begin{align}
x_{n+1} &= f(x_n) \quad (\$) \\
x_{n+1} &= g(x_n)
\end{align}
The fixed point iteration of $f$ is like the the iteration of original continued fractions (compare $(\$)$ with $(*)$ or $(**)$).
The theory behind can now help with statements about convergence and the dependency of start values.
A: Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$.
$$
\begin{align}
n&=1&
3&=3&
3-2&=1\\\\
n&=2&
3-\cfrac23&=\frac73&
3-\cfrac{2}{3-2}&=1\\\\
n&=3&
3-\cfrac{2}{3-\cfrac23}&=\frac{15}7&
3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\
n&=4&
3-\cfrac2{3-\cfrac2{3-\cfrac23}}&=\frac{31}{15}&
3-\cfrac2{3-\cfrac2{3-\cfrac2{3-2}}}&=1\\\\
&&
3-\cfrac2{\cfrac{2^n-1}{2^{n-1}-1}}&=\frac{2^{n+1}-1}{2^n-1}&
3-\cfrac21&=1
\end{align}
$$
Therefore, the limit of the continued fractions with $2n-1$ twos and threes is
$$
\lim_{n\to\infty}\frac{2^{n+1}-1}{2^n-1}=2
$$
and the limit of the continued fractions with $2n$ twos and threes is
$$
\lim_{n\to\infty}1=1
$$
Therefore, one value represents the limit of an odd number of twos and threes and the other value represents the limit of an even number of twos and threes.
A: Your two solutions are correct. You can verify them as follows.
$$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$
Now replace the red 1 with the blue 1, which equals to the right hand side in (1).
$$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$
Now replace the red 2 with the blue 2, which equals to the right hand side in (2).
I heard from the internet that this is one of the ways Ramanujan created some of his equalities.
Now 2 questions for you, 
(1) what if you replace the cyan 2 with blue 2 and so on so forth?
(2) what if you replace the cyan 2 with blue 2 and then replace the red 2 with blue 2 and so on so forth?
