parmenides51 wrote: Find out which of the two numbers is greater: $$\dfrac{2}{2 +\dfrac{2}{2 +\dfrac{2}{... +\dfrac{2}{2+\frac22}}}} \,\,\, \text{or} \,\,\, \dfrac{3}{3 +\dfrac{3}{3 +\dfrac{3}{... +\dfrac{3}{3+\frac33}}}}$$(Each expression has $2022$ fraction signs.) Let $u_n=\frac 2{2+\frac 2{...}}$ and $v_n=\frac 3{3+\frac 3{...}}$ with $ n$ fractions signs If $u_n<v_n$, then $2+u_n<3+v_n$ and so $1+\frac 1{2+u_n}>1+\frac 1{3+v_n}$, and so $\frac 1{1+\frac 1{2+u_n}}<\frac 1{1+\frac 1{3+v_n}}$ Which is $\frac 2{2+\frac 2{2+u_n}}<\frac 3{3+\frac 3{3+v_n}}$, which is $u_{n+2}<v_{n+2}$ $u_2=\frac 2{2+\frac 22}=\frac 23<\frac 34=\frac 3{3+\frac 33}=v_2$ And so $u_{2n}<v_{2n}$

Surely you can say that if $x=\frac{2}{2+x}$, then $x^2+2x-2=0$ so $x= -1+\sqrt{5}<1.24$, while for the latter $x=\frac{3}{3+x}$ gives $x^2+3x-3=0$ so we have $x=\frac{3+\sqrt{21}}{2}>3.79$. This is not a proof, but with extremely weak bounds on the rate of converge of these continued fractions, you you can make this rigorous.