I attempt to prove that ${\pi}$ is an irrational number. For this, I use the beautiful continued fraction given by Brouncker who rewrote Wallis' formula as a continued fraction, which Wallis and later Euler (1775) proved to be equivalent.
The continued fraction given by Brouncker is
$$\frac{4}{\pi} =
1 + \cfrac{1}{3 + \cfrac{4}{5 + \cfrac{9}{7 + \dots}}}$$
Now we know that every infinite continued fraction is irrational. Does it mean that since RHS has an infinite continued fraction the LHS $\frac{4}{\pi}$ must also be irrational? And hence ${\pi}$ must be an irrational number?
EDIT 1: From this post General Continued Fractions and Irrationality If, in the continued fraction
$$\cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \dots}}}$$
the values $a_{i}, b_{i}$ are all positive integers, and if we have $a_i \geq b_i$ for all $i$ greater than some $n$, then the value of the continued fraction is irrational(Is this a necessary and sufficient condition? I am in doubt since the golder relation $\phi$ has $a_i \geq b_i=1$ for all $n\geq1$). As the condition is not satisfied the proof is actually wrong.
EDIT 2:
How to prove the following continued fraction of $e^{x/y}$
$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$
Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $e^{x/y}$ will be an irrational number!
Best Answer
To see that not every generalized infinite continued fraction is irrational, note that if $$ x=1 + \frac{2}{1 + \frac{2}{1 + \frac{2}{1 + ...}}} $$ then $x=1+{\large{\frac{2}{x}}}$, so $x=2$.
Of course that assumes that the continued fraction converges.
To prove $x=2$ more rigorously, we can argue as follows . . .
By definition, $x$ is the limit, if it exists, of the infinite sequence $x_0,x_1,x_2,...$ defined recursively by $x_0=1$ and $$ x_n=1+\frac{2}{x_{n-1}} $$ for $n\ge 1$.
Clearly we have $x_n \ge 1$ for all $n$.
For $n\ge 1$ we have $$ x_n=1+\frac{2}{x_{n-1}}\le 1+\frac{2}{1}=3 $$ hence, since $x_0=1$, we have $x_n\le 3$ for all $n$.
Then for all $n\ge 1$ we get $$ x_n=1+\frac{2}{x_{n-1}}\ge 1+\frac{2}{3}=\frac{5}{3} $$ From the recursion we get $$ x_n-2=\frac{2-x_{n-1}}{x_{n-1}} $$ for all $n\ge 1$, hence for all $n\ge 2$ we get $$ \left|x_{n}-2\right|\le\frac{\left|x_{n-1}-2\right|}{\left({\large{\frac{5}{3}}}\right)} $$ and then an easy induction yields $$ \left|x_n-2\right|\le\Bigl(\frac{3}{5}\Bigr)^{\large{n-1}} $$ for all $n\ge 2$.
It follows that $$ \lim_{n\to\infty}x_n = 2 $$ so $x=2$.
As to the question you raised in your edit, consider the generalized continued fraction $$ 1 + \frac{3}{1 + \frac{3}{1 + \frac{3}{1 + ...}}} $$ which can be shown to be equal to $$ \frac{1+\sqrt{13}}{2} $$ Thus the condition for irrationality that you referenced is a sufficient condition, but not a necessary condition.